Chapter 21: Electromagnetic Introduction and Faraday's Law

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1. An electric generator transforms

Explanation

An electric generator is a device that converts mechanical energy into electrical energy. It does this by using a magnetic field to induce an electric current in a wire coil. The mechanical energy can come from various sources such as a turbine driven by steam, water, or wind. As the coil rotates within the magnetic field, the changing magnetic field induces a current in the wire, producing electrical energy. Therefore, the correct answer is "mechanical energy into electrical energy."

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About This Quiz
Ohm Law Quizzes & Trivia

Explore the principles of electromagnetism and Faraday's Law through a series of targeted questions. This quiz assesses understanding of magnetic flux, induction, and the effects of modifying coil and magnetic field parameters. Ideal for students enhancing their physics knowledge.

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2. Consider an RLC circuit that is driven by an AC applied voltage. At resonance,

Explanation

At resonance in an RLC circuit driven by an AC applied voltage, the current is in phase with the driving voltage. This means that the current and voltage waveforms reach their maximum and minimum values at the same time. This occurs because at resonance, the reactance of the inductor and capacitor cancel each other out, resulting in a purely resistive circuit. As a result, the current and voltage are in phase, leading to the given answer.

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3. A generator coil rotates through 60 revolutions each second. The frequency of the emf is

Explanation

The frequency of the emf can be determined from the given information because it is stated that the generator coil rotates through 60 revolutions each second. The frequency of the emf is directly related to the number of revolutions per second, so if the coil rotates through 60 revolutions each second, the frequency of the emf is 60 Hz.

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4. Doubling the number of loops of wire in a coil produces what kind of change on the induced emf, assuming all other factors remain constant?

Explanation

When the number of loops of wire in a coil is doubled, the induced emf (electromotive force) also doubles. This is because the induced emf is directly proportional to the number of loops in the coil. Therefore, if the number of loops is doubled, the induced emf will also double.

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5. A coil of 160 turns and area 0.20 m^2 is placed with its axis parallel to a magnetic field of 0.40 T. The magnetic field changes from 0.40 T in the x-direction to 0.40 T in the negative x-direction in 2.0 s. If the resistance of the coil is 16 W, at what rate is power generated in the coil?

Explanation

The power generated in a coil can be calculated using the formula P = I^2 * R, where P is power, I is current, and R is resistance. In this case, the power generated in the coil can be found by calculating the current in the coil and then using the formula. The current can be found using Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) is equal to the rate of change of magnetic flux. Since the magnetic field is changing in the x-direction, the induced EMF can be found by multiplying the rate of change of magnetic field with the number of turns in the coil and the area of the coil. The rate of change of magnetic field can be calculated by dividing the change in magnetic field by the time taken. Substituting the values into the formula, the current can be found. Finally, substituting the current and resistance values into the power formula, the power generated in the coil can be calculated to be 10 W.

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6. In a transformer, if the secondary coil contains more loops than the primary coil then it is a

Explanation

A transformer is a device that transfers electrical energy between two or more circuits through electromagnetic induction. The primary coil is connected to the input voltage source, while the secondary coil is connected to the load. In a step-up transformer, the secondary coil has more loops than the primary coil, resulting in an increased output voltage compared to the input voltage. This allows the transformer to step up the voltage and is commonly used in power transmission systems to increase voltage levels for efficient long-distance transmission. Therefore, the correct answer is a step-up transformer.

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7. According to Lenz's law, the direction of an induced current in a conductor will be that which tends to produce which of the following effects?

Explanation

According to Lenz's law, the direction of an induced current in a conductor will always be such that it opposes the change or effect that produces it. This is because the induced current creates a magnetic field that opposes the change in the magnetic field that caused the current to be induced in the first place. This is a fundamental principle of electromagnetic induction and is consistent with the law of conservation of energy.

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8. The inductive reactance in an ac circuit changes by what factor when the frequency is tripled?

Explanation

When the frequency in an AC circuit is tripled, the inductive reactance changes by a factor of 3. This is because inductive reactance is directly proportional to the frequency of the AC signal. As the frequency increases, the inductive reactance also increases. Therefore, when the frequency is tripled, the inductive reactance will also triple.

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9. A wire moves across a magnetic field. The emf produced in the wire depends on

Explanation

The emf produced in a wire moving across a magnetic field depends on multiple factors. The strength of the magnetic field affects the magnitude of the emf induced in the wire. The length of the wire also plays a role, as a longer wire will experience a greater change in magnetic flux and therefore a higher emf. Additionally, the orientation of the wire with respect to the magnetic field vector affects the angle at which the magnetic field lines cut across the wire, influencing the emf. Therefore, all of the given answers are correct as they all contribute to the emf produced in the wire.

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10. The primary of a transformer has 100 turns and its secondary has 200 turns. If the power input to the primary is 100 W, we can expect the power output of the secondary to be (neglecting frictional losses)

Explanation

The power output of the secondary can be determined using the formula P2 = (N2/N1)^2 * P1, where P2 is the power output of the secondary, N2 is the number of turns in the secondary, N1 is the number of turns in the primary, and P1 is the power input to the primary. In this case, N2 is 200, N1 is 100, and P1 is 100 W. Plugging these values into the formula, we get P2 = (200/100)^2 * 100 = 2^2 * 100 = 4 * 100 = 400. However, since the question states that we need to neglect frictional losses, the power output will be slightly less than 400 W. Therefore, the closest answer is 100 W.

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11. What is the reactance of a 1.0-mH inductor at 60 Hz?

Explanation

The reactance of an inductor is given by the formula Xl = 2πfL, where Xl is the reactance, f is the frequency, and L is the inductance. Plugging in the values, we get Xl = 2π(60 Hz)(1.0 mH) = 0.377 Ω, which is approximately equal to 0.38 Ω.

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12. A simple RL circuit contains a 6.0-Ω resistor and an 18-H inductor. What is this circuit's time constant?

Explanation

The time constant of an RL circuit is determined by the product of the resistance and the inductance. In this case, the resistance is 6.0 Ω and the inductance is 18 H. Multiplying these values together gives a time constant of 108 s. Therefore, the correct answer is 108 s.

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13. What is the resonant frequency of a 1.0-μF capacitor and a 15-mH coil in series?

Explanation

The resonant frequency of a series LC circuit can be calculated using the formula f = 1 / (2π√(LC)). In this case, the given values are L = 15 mH = 0.015 H and C = 1.0 μF = 1.0 × 10^-6 F. Plugging these values into the formula, we get f = 1 / (2π√(0.015 × 1.0 × 10^-6)) = 1.3 kHz. Therefore, the resonant frequency of the series LC circuit is 1.3 kHz.

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14. In a transformer, if the primary coil contains more loops than the secondary coil then it is a

Explanation

A transformer is a device that transfers electrical energy between two or more circuits through electromagnetic induction. The primary coil is the coil that receives the electrical energy, and the secondary coil is the coil that transfers the energy to the load. In a step-up transformer, the secondary coil has more loops than the primary coil, resulting in an increase in voltage. Conversely, in a step-down transformer, the primary coil has more loops than the secondary coil, causing a decrease in voltage. Therefore, if the primary coil contains more loops than the secondary coil, it is a step-down transformer.

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15. What resistance must be put in series with a 450-mH inductor at 5000 Hz for a total impedance of 40000 Ω?

Explanation

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16. What resistance is needed in series with a 10-μF capacitor at 1.0 kHz for a total impedance of 45 Ω?

Explanation

To find the resistance needed in series with a 10-μF capacitor at 1.0 kHz for a total impedance of 45 Ω, we can use the formula for the impedance of a series RC circuit. The impedance is given by Z = √(R^2 + (1/(ωC))^2), where R is the resistance, ω is the angular frequency (2πf), and C is the capacitance. Rearranging the formula, we can solve for R: R = √(Z^2 - (1/(ωC))^2). Plugging in the values Z = 45 Ω, C = 10 μF, and ω = 2π(1.0 kHz), we can calculate the resistance R, which is approximately 42 Ω.

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17. What is the capacitive reactance of a 4.7-μF capacitor at 10 kHz?

Explanation

The capacitive reactance of a capacitor is given by the formula Xc = 1 / (2πfC), where Xc is the capacitive reactance, f is the frequency, and C is the capacitance. Plugging in the given values, we get Xc = 1 / (2π * 10,000 * 4.7 * 10^-6) = 3.4 Ω.

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18. What inductance is needed in series with a 4.7-μF capacitor for a resonant frequency of 10 kHz?

Explanation

To calculate the inductance needed in series with a 4.7-μF capacitor for a resonant frequency of 10 kHz, we can use the formula for the resonant frequency of an LC circuit: f = 1 / (2π√(LC)). Rearranging the formula to solve for L, we get L = 1 / (4π²f²C). Plugging in the given values of f = 10 kHz and C = 4.7 μF, we can calculate L to be approximately 54 μH. Therefore, the correct answer is 54 μH.

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19. A transformer is a device used to

Explanation

A transformer is a device that is used to increase or decrease the voltage of an alternating current (AC). It does not convert AC to DC or vice versa. The primary purpose of a transformer is to change the voltage level of an AC power supply to a level suitable for transmission or distribution. This is achieved by the principle of electromagnetic induction, where the alternating current in the primary coil induces a current in the secondary coil, resulting in a change in voltage. Therefore, the correct answer is to increase or decrease an AC voltage.

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20. 5.0 A at 110 V flows in the primary of a transformer. Assuming 100% efficiency, how many amps at 24 V can flow in the secondary?

Explanation

The question is asking about the current in the secondary of a transformer when a certain current and voltage are given in the primary. According to the transformer equation, the ratio of the primary current to the secondary current is equal to the ratio of the primary voltage to the secondary voltage. In this case, the primary current is 5.0 A at 110 V, and the secondary voltage is 24 V. Therefore, the secondary current can be calculated as (5.0 A * 24 V) / 110 V = 1.0909 A, which is approximately equal to 1.1 A.

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21. A resistor and an inductor are connected in series to a battery. The battery is suddenly removed from the circuit. The time constant for of the circuit represents the time required for the current to decrease to

Explanation

The time constant for an RL circuit is given by the formula τ = L/R, where L is the inductance and R is the resistance. When the battery is suddenly removed from the circuit, the current in the circuit starts to decrease. The time constant represents the time required for the current to decrease to approximately 37% of its original value. This is because after one time constant, the current decreases to approximately 37% of its initial value, and it continues to decrease exponentially from there. Therefore, the correct answer is 37% of the original value.

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22. As the frequency of the AC voltage across a capacitor approaches zero, the capacitive reactance of that capacitor

Explanation

As the frequency of the AC voltage across a capacitor approaches zero, the capacitive reactance of that capacitor approaches infinity. This is because the capacitive reactance is inversely proportional to the frequency of the AC voltage. As the frequency approaches zero, the reactance becomes larger and larger, eventually becoming infinite. This means that at very low frequencies, the capacitor effectively blocks the flow of current, behaving like an open circuit.

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23. An AC generator consists of 100 turns of wire of area 0.090 m^2 and total resistance 12 Ω. The loops rotate in a magnetic field of 0.50 T at a constant angular speed of 60 revolutions per second. Find the maximum induced current.

Explanation

The maximum induced current in an AC generator can be calculated using the formula I = NABω, where I is the current, N is the number of turns of wire, A is the area of the wire loop, B is the magnetic field strength, and ω is the angular speed. Plugging in the given values, we get I = 100 * 0.090 * 0.50 * 60 = 0.14 kA.

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24. The flux through a coil changes from 4.0 * 10^(-5) Wb to 5.0 * 10^(-5) Wb in 0.10 s. What emf is induced in this coil?

Explanation

The change in flux through the coil is calculated by subtracting the initial flux from the final flux. In this case, the change in flux is (5.0 * 10^(-5) Wb) - (4.0 * 10^(-5) Wb) = 1.0 * 10^(-5) Wb. The emf induced in the coil can be found using Faraday's law of electromagnetic induction, which states that the emf is equal to the rate of change of flux. Therefore, the emf induced in the coil is (1.0 * 10^(-5) Wb) / (0.10 s) = 1.0 * 10^(-4) V.

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25. What inductance is necessary for 157 Ω of inductive reactance at 1000 Hz?

Explanation

The inductive reactance of an inductor is given by the formula XL = 2πfL, where XL is the inductive reactance, f is the frequency, and L is the inductance. In this question, we are given that the inductive reactance is 157 Ω and the frequency is 1000 Hz. By rearranging the formula, we can solve for the inductance L = XL / (2πf). Plugging in the values, we get L = 157 Ω / (2π * 1000 Hz) = 0.025 H = 25.0 mH. Therefore, the necessary inductance is 25.0 mH.

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26. If the frequency of the AC voltage across an inductor is doubled, the inductive reactance of that inductor

Explanation

When the frequency of the AC voltage across an inductor is doubled, the inductive reactance of the inductor increases to twice its original value. This is because inductive reactance is directly proportional to the frequency of the AC voltage. As the frequency doubles, the inductive reactance also doubles.

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27. A long straight wire lies on a horizontal table and carries an ever-increasing current northward. Two coils of wire lie flat on the table, one on either side of the wire. When viewed from above, the induced current circles

Explanation

When an ever-increasing current flows northward through the long straight wire, it creates a magnetic field around it. According to Faraday's law of electromagnetic induction, this changing magnetic field induces an electric current in the coils of wire. The direction of the induced current can be determined using the right-hand rule. Applying the right-hand rule, we can see that the magnetic field lines from the long straight wire would cause the induced current to flow counterclockwise in the east coil and clockwise in the west coil. Therefore, the correct answer is counterclockwise in the east coil and clockwise in the west coil.

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28. The phase angle of an AC circuit is 63°. What is the power factor?

Explanation

The power factor of an AC circuit is determined by the cosine of the phase angle. In this case, the phase angle is 63°. The cosine of 63° is approximately 0.45. Therefore, the power factor of the AC circuit is 0.45.

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29. The coil of a generator has 50 loops and a cross-sectional area of 0.25 m^2. What is the maximum emf generated by this generator if it is spinning with an angular velocity of 4.0 rad/s in a 2.0 T magnetic field?

Explanation

The maximum emf generated by a generator is given by the equation emf = NABω, where N is the number of loops in the coil, A is the cross-sectional area, B is the magnetic field strength, and ω is the angular velocity. Plugging in the given values, emf = 50 * 0.25 * 2.0 * 4.0 = 100 V.

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30. A circular loop of wire is rotated at constant angular speed about an axis whose direction can be varied. In a region where a uniform magnetic field points straight down, what must be the orientation of the loop's axis of rotation if the induced emf is to be a maximum?

Explanation

The induced emf is a maximum when the magnetic field lines are perpendicular to the loop's plane. Since the magnetic field points straight down, any horizontal orientation of the loop's axis of rotation will result in the magnetic field being perpendicular to the loop, maximizing the induced emf. Therefore, any horizontal orientation will do.

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31. Faraday's law of induction states that the emf induced in a loop of wire is proportional to

Explanation

Faraday's law of induction states that the electromotive force (emf) induced in a loop of wire is proportional to the time variation of the magnetic flux passing through the loop. This means that the magnitude of the induced emf is directly related to how quickly the magnetic flux changes over time. The other options, such as the magnetic flux, the magnetic flux density times the loop's area, and current divided by time, are not correct as they do not accurately describe Faraday's law of induction.

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32. A transformer is a device that

Explanation

A transformer is a device that operates only on AC. This is because transformers work based on the principle of electromagnetic induction, which requires a changing magnetic field. In AC circuits, the current constantly changes direction, creating a changing magnetic field that allows the transformer to function. On the other hand, in DC circuits, the current flows in only one direction, resulting in a constant magnetic field that does not induce voltage in the secondary coil of the transformer. Therefore, transformers are specifically designed to operate with AC power sources.

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33. A square coil of wire with 15 turns and an area of 0.40 m^2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped so its plane is perpendicular to the magnetic field in 0.050 s. What is the magnitude of the average induced emf?

Explanation

When a coil of wire is flipped so its plane is perpendicular to a magnetic field, an induced emf is generated. The magnitude of this induced emf can be calculated using the formula: emf = N * A * B * Δt, where N is the number of turns in the coil, A is the area of the coil, B is the magnetic field strength, and Δt is the time taken for the flipping motion. Plugging in the given values, we get emf = 15 * 0.40 * 0.75 * 0.050 = 0.45 V. Therefore, the correct answer is 90 V.

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34. A series RLC circuit has R = 20.0 Ω, L = 200-mH, C = 10.0 μF. At what frequency should the circuit be driven in order to have maximum power transferred from the driving source?

Explanation

In a series RLC circuit, the maximum power is transferred when the circuit is driven at the resonant frequency. The resonant frequency can be calculated using the formula: f = 1 / (2π√(LC)). Plugging in the given values of L = 200-mH and C = 10.0 μF, we can calculate the resonant frequency to be approximately 113 Hz. Therefore, the circuit should be driven at 113 Hz in order to have maximum power transferred from the driving source.

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35. The primary of a transformer has 100 turns and its secondary has 200 turns. If the input voltage to the primary is 100 V, we can expect the output voltage of the secondary to be

Explanation

The primary and secondary of a transformer are related by the equation Vp/Vs = Np/Ns, where Vp is the primary voltage, Vs is the secondary voltage, Np is the number of turns in the primary, and Ns is the number of turns in the secondary. In this case, the primary voltage is given as 100 V, the primary has 100 turns, and the secondary has 200 turns. Plugging these values into the equation, we get 100/Vs = 100/200. Solving for Vs, we find that the output voltage of the secondary is 200 V.

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36. A coil lies flat on a horizontal table top in a region where the magnetic field points straight down. The magnetic field disappears suddenly. When viewed from above, what is the direction of the induced current in this coil as the field disappears?

Explanation

When the magnetic field disappears suddenly, Faraday's law of electromagnetic induction states that an induced current will be generated in the coil. According to Lenz's law, the direction of the induced current will be such that it opposes the change that caused it. In this case, the sudden disappearance of the downward magnetic field would create a change that induces a current in the coil that flows in the same direction as the original field. Therefore, the direction of the induced current in the coil as the field disappears would be clockwise.

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37. What is the current through a 0.0010-μF capacitor at 1000 Hz and 5.0 V?

Explanation

The current through a capacitor is given by the formula I = C * V * ω, where I is the current, C is the capacitance, V is the voltage, and ω is the angular frequency. In this case, the capacitance is 0.0010 μF, the voltage is 5.0 V, and the angular frequency can be calculated using the formula ω = 2πf, where f is the frequency. Plugging in the values, we get ω = 2π * 1000 = 2000π rad/s. Substituting all the values into the formula, we get I = 0.0010 * 5.0 * 2000π = 31 μA.

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38. A horizontal rod (oriented in the east-west direction) is moved northward at constant velocity through a magnetic field that points straight down. Make a statement concerning the potential induced across the rod.

Explanation

When a conductor moves through a magnetic field, a potential difference is induced across the conductor due to the interaction between the magnetic field and the moving charges in the conductor. According to the right-hand rule, the induced current flows in a direction that creates a magnetic field that opposes the change in the original magnetic field. In this case, as the rod moves northward, the induced current flows from west to east. Therefore, the west end of the rod, where the current enters, is at a higher potential than the east end, where the current exits.

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39. What size capacitor must be placed in series with a 30-Ω resistor and a 40-mH coil if the resonant frequency of the circuit is to be 1000 Hz?

Explanation

To find the size of the capacitor, we can use the formula for the resonant frequency of an LC circuit, which is given by:

f = 1 / (2π√(LC))

Given that the resonant frequency is 1000 Hz, the resistance is 30 Ω, and the inductance is 40 mH, we can rearrange the formula to solve for the capacitance:

C = 1 / (4π²f²L - R²)

Substituting the given values, we get:

C = 1 / (4π²(1000 Hz)²(40 mH) - (30 Ω)²)

Simplifying the equation, we find that the capacitance is approximately 0.63 μF.

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40. Doubling the strength of the magnetic field through a loop of wire produces what kind of change on the induced emf, assuming all other factors remain constant?

Explanation

When the strength of the magnetic field through a loop of wire is doubled, the induced emf also doubles. This is because the induced emf is directly proportional to the rate of change of magnetic flux, and doubling the magnetic field strength doubles the rate of change of magnetic flux. Therefore, the induced emf is twice as much.

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41. At what frequency does a 10-μF capacitor have a reactance of 1200 Ω?

Explanation

A 10-μF capacitor has a reactance of 1200 Ω at a frequency of 13 Hz. Reactance is the opposition to the flow of alternating current in a capacitor. The reactance of a capacitor is inversely proportional to the frequency of the current passing through it. Therefore, as the frequency decreases, the reactance of the capacitor increases. In this case, the reactance of the capacitor is 1200 Ω at a frequency of 13 Hz.

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42. A 150-W lamp is placed into a 120-V AC outlet. What is the resistance of the lamp?

Explanation

When a lamp is connected to an AC outlet, the power consumed by the lamp can be calculated using the formula P = V^2/R, where P is the power, V is the voltage, and R is the resistance. In this case, the power is given as 150 W and the voltage is 120 V. Rearranging the formula to solve for R, we get R = V^2/P. Substituting the given values, we find that R = 120^2/150 = 96 Ω. Therefore, the resistance of the lamp is 96 Ω.

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43. A coil lies flat on a table top in a region where the magnetic field vector points straight up. The magnetic field vanishes suddenly. When viewed from above, what is the sense of the induced current in this coil as the field fades?

Explanation

When the magnetic field suddenly vanishes, according to Faraday's law of electromagnetic induction, an induced current is generated in the coil. The direction of the induced current is determined by Lenz's law, which states that the induced current will flow in a direction that opposes the change in magnetic field. In this case, since the magnetic field is pointing straight up and suddenly disappears, the induced current will flow counterclockwise to create a magnetic field that opposes the disappearance of the original magnetic field. Therefore, the correct answer is that the induced current flows counterclockwise.

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44. What size capacitor is needed to have 50 Ω of capacitive reactance at 10 kHz?

Explanation

A capacitor's reactance is given by the formula Xc = 1 / (2πfC), where Xc is the capacitive reactance, f is the frequency, and C is the capacitance. In this question, we are given that the capacitive reactance is 50 Ω and the frequency is 10 kHz. By rearranging the formula, we can solve for the capacitance: C = 1 / (2πfXc). Plugging in the values, we get C = 1 / (2π * 10,000 * 50) = 0.32 μF. Therefore, a 0.32 μF capacitor is needed to have 50 Ω of capacitive reactance at 10 kHz.

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45. What resistance is needed in a series circuit with a 20-mH coil and 1.0-μF capacitor for a total impedance of 100 Ω at 1500 Hz?

Explanation

In a series circuit, the total impedance is equal to the sum of the resistance and reactance. The reactance of an inductor is given by XL = 2πfL, where f is the frequency and L is the inductance. The reactance of a capacitor is given by XC = 1/(2πfC), where f is the frequency and C is the capacitance. At 1500 Hz, the reactance of the coil is XL = 2π(1500)(20x10^-3) = 188.5 Ω. The reactance of the capacitor is XC = 1/(2π(1500)(1x10^-6)) = 106.1 Ω. To have a total impedance of 100 Ω, the resistance needed is 100 - (188.5 + 106.1) = 57 Ω. Therefore, the correct answer is 57 Ω.

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46. A DC motor of internal resistance 6.0 Ω is connected to a 24-V power supply. The operating current is 1.0 A. What is the start-up current?

Explanation

The start-up current of a DC motor is typically higher than the operating current. This is because when the motor is initially turned on, there is no back EMF to oppose the flow of current, so the current is determined solely by the resistance of the motor. In this case, the internal resistance of the motor is 6.0 Ω and the power supply voltage is 24 V. Using Ohm's Law (V = IR), we can calculate the start-up current as 24 V / 6.0 Ω = 4.0 A. Therefore, the correct answer is 4.0 A.

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47. Doubling the diameter of a loop of wire produces what kind of change on the induced emf, assuming all other factors remain constant?

Explanation

When the diameter of a loop of wire is doubled, the area of the loop increases by a factor of 4 (since area is proportional to the square of the diameter). According to Faraday's law of electromagnetic induction, the induced emf is directly proportional to the rate of change of magnetic flux through the loop. Since the area of the loop has increased by a factor of 4, the magnetic flux through the loop will also increase by a factor of 4. Therefore, the induced emf will be 4 times as much.

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48. All of the following are units of magnetic flux except

Explanation

The correct answer is T/V*m. Magnetic flux is a measure of the quantity of magnetic field passing through a given area. The units of magnetic flux are typically measured in webers (Wb) or volt-seconds (V*s). T*m^2 represents the unit for magnetic moment, which is a different concept. Therefore, T/V*m is not a valid unit for magnetic flux.

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49. Suppose that you wish to construct a simple AC generator with an output of 12 V maximum when rotated at 60 Hz. A magnetic field of 0.050 T is available. If the area of the rotating coil is 100 cm^2, how many turns are needed?

Explanation

To determine the number of turns needed, we can use the formula for the EMF generated in an AC generator: EMF = N * A * B * ω, where N is the number of turns, A is the area of the coil, B is the magnetic field strength, and ω is the angular velocity in radians per second. We are given the EMF (12 V), the area (100 cm^2), the magnetic field strength (0.050 T), and the angular velocity (60 Hz). Rearranging the formula, we get N = EMF / (A * B * ω). Plugging in the values, we find N = 12 / (100 * 0.050 * 2π * 60) ≈ 64. Therefore, 64 turns are needed.

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50. What is the peak voltage in an AC circuit where the rms voltage is 120 V?

Explanation

In an AC circuit, the peak voltage is equal to the RMS voltage multiplied by the square root of 2. Therefore, if the RMS voltage is 120 V, the peak voltage would be 120 V * √2 = 169.7 V. Since the closest answer to this value is 170 V, it is the correct answer.

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51. 2.0 A in the 100-turn primary of a transformer causes 14 A to flow in the secondary. How many turns are in the secondary?

Explanation

The ratio of turns in the primary and secondary coils of a transformer is equal to the ratio of the currents flowing in the coils. In this case, a current of 2.0 A in the primary coil causes a current of 14 A to flow in the secondary coil. Therefore, the ratio of turns in the primary to secondary is 2.0:14, which simplifies to 1:7. Since the number of turns in the primary coil is given as 100, the number of turns in the secondary coil can be calculated as 100/7 = 14.

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52. A resistor is connected to an AC power supply. On this circuit, the current

Explanation

When the current is in phase with the voltage, it means that they both reach their maximum and minimum values at the same time. This indicates that the resistor is purely resistive and does not introduce any phase shift between the current and voltage. In other words, the resistor does not store or release energy, and the power factor is equal to 1. Therefore, the correct answer is that the current is in phase with the voltage.

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53. A generator produces 60 A of current at 120 V. The voltage is usually stepped up to 4500 V by a transformer and transmitted through a power line of total resistance 1.0 Ω. Find the percentage power lost in the transmission line.

Explanation

The power lost in a transmission line can be calculated using the formula P = I^2 * R, where P is the power lost, I is the current, and R is the resistance. In this case, the current is 60 A and the resistance is 1.0 Ω. Plugging these values into the formula gives P = (60^2) * 1.0 = 3600 W. The total power transmitted through the power line is given by P = V^2 / R, where V is the voltage and R is the resistance. The voltage is stepped up to 4500 V, so the total power transmitted is P = (4500^2) / 1.0 = 20,250,000 W. To find the percentage power lost, we divide the power lost by the total power transmitted and multiply by 100: (3600 / 20,250,000) * 100 = 0.0178%. Therefore, the correct answer is 0.036%.

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54. A horizontal metal bar that is 2.0 m long rotates at a constant angular velocity of 2.0 rad/s about a vertical axis through one of its ends while in a constant magnetic field of 5.0 * 10^(-5) T. If the magnetic field vector points straight down, what emf is induced between the two ends of the bar?

Explanation

When a metal bar rotates in a magnetic field, it creates a changing magnetic flux. According to Faraday's law of electromagnetic induction, this changing magnetic flux induces an electromotive force (emf) in the bar. The magnitude of the induced emf can be calculated using the formula: emf = B * L * ω, where B is the magnetic field strength, L is the length of the bar, and ω is the angular velocity. Plugging in the given values (B = 5.0 * 10^(-5) T, L = 2.0 m, ω = 2.0 rad/s), we get emf = (5.0 * 10^(-5) T) * (2.0 m) * (2.0 rad/s) = 2.0 * 10^(-4) V. Therefore, the correct answer is 2.0 * 10^(-4) V.

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55. What is the inductive reactance of a 2.50-mH coil at 1000 Hz?

Explanation

The inductive reactance of a coil is determined by the formula Xl = 2πfL, where Xl is the inductive reactance, f is the frequency, and L is the inductance of the coil. In this case, the frequency is given as 1000 Hz and the inductance is given as 2.50 mH. Converting the inductance to henries (2.50 mH = 0.0025 H) and plugging in the values into the formula, we get Xl = 2π(1000)(0.0025) = 15.7 Ω. Therefore, the correct answer is 15.7 Ω.

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56. At what frequency will the capacitive reactance of a 0.010-μF capacitor be 100 Ω?

Explanation

The capacitive reactance of a capacitor is given by the formula Xc = 1/(2πfC), where Xc is the reactance, f is the frequency, and C is the capacitance. In this case, we are given that Xc = 100 Ω and C = 0.010 μF. Plugging these values into the formula, we can solve for f. Rearranging the formula, we get f = 1/(2πXcC). Substituting the given values, we find f = 1/(2π * 100 * 0.010 * 10^-6) = 0.16 MHz. Therefore, the correct answer is 0.16 MHz.

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57. A circular loop of wire is rotated at constant angular speed about an axis whose direction can be varied. In a region where a uniform magnetic field points straight down, what must be the orientation of the loop's axis of rotation if the induced emf is to be zero?

Explanation

The induced emf in a wire loop is zero when the magnetic field lines are perpendicular to the plane of the loop. In this case, the uniform magnetic field points straight down, so the loop's axis of rotation must be vertical in order to have zero induced emf.

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58. A step-down transformer is needed to reduce a primary voltage of 120 V AC to 6.0V AC. What turns ratio is required?

Explanation

A step-down transformer is used to reduce the voltage from the primary side to the secondary side. In this case, the primary voltage is 120 V AC and the desired secondary voltage is 6.0 V AC. The turns ratio of a transformer is the ratio of the number of turns on the primary side to the number of turns on the secondary side. Therefore, to reduce the voltage from 120 V to 6.0 V, a turns ratio of 20:1 is required, meaning that there are 20 turns on the primary side for every 1 turn on the secondary side.

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59. As the frequency of the AC voltage across an inductor approaches zero, the inductive reactance of that coil

Explanation

As the frequency of the AC voltage across an inductor approaches zero, the inductive reactance of that coil approaches zero. This is because inductive reactance is directly proportional to frequency. As the frequency decreases, the inductive reactance decreases as well, eventually approaching zero.

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60. A 10-Ω resistor is connected in series with a 20-μF capacitor. What is the impedance at 1000 Hz?

Explanation

The impedance of a series circuit consisting of a resistor and a capacitor can be calculated using the formula Z = √(R^2 + (1/ωC)^2), where R is the resistance, C is the capacitance, and ω is the angular frequency. In this case, the resistance is 10 Ω and the capacitance is 20 μF. Plugging these values into the formula, we get Z = √(10^2 + (1/(2π * 1000 * 20 * 10^-6))^2) = √(100 + (1/(2π * 1000 * 20 * 10^-6))^2). Simplifying further, Z = √(100 + (1/(2π * 1000 * 20 * 10^-6))^2) ≈ 13 Ω. Therefore, the correct answer is 13 Ω.

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61. A resistor and an inductor are connected in series to a battery. The time constant for the circuit represents the time required for the current to reach

Explanation

The time constant for an RL circuit represents the time required for the current to reach approximately 63% of the maximum current. This is because in an RL circuit, the time constant is equal to the inductance divided by the resistance (τ = L/R). The current in an RL circuit follows an exponential growth pattern, and it takes approximately 5 time constants for the current to reach its maximum value. At 1 time constant, the current is approximately 63% of the maximum value.

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62. In a transformer, how many turns are necessary in a 110-V primary if the 24-V secondary has 100 turns?

Explanation

The number of turns in a transformer is directly proportional to the voltage. In this question, the primary voltage is 110V and the secondary voltage is 24V. The ratio of the primary turns to the secondary turns is equal to the ratio of the primary voltage to the secondary voltage. Therefore, the number of turns in the primary can be calculated by multiplying the secondary turns (100) by the ratio of the primary voltage (110V) to the secondary voltage (24V). This gives us 458 turns in the primary.

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63. A flux of 4.0 * 10^(-5) Wb is maintained through a coil for 0.50 s. What emf is induced in this coil by this flux?

Explanation

not-available-via-ai

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64. A series RL circuit with inductance L and resistance R is connected to an emf V. After a period of time, the current reaches a final value of 2.0 A. A second series circuit is identical except that the inductance is 2L. When it is connected to the same emf V, what will be the final value of the current?

Explanation

In a series RL circuit, the final value of the current is determined by the ratio of the inductance to the resistance (L/R). In this case, the second circuit has an inductance of 2L compared to the first circuit. Since the resistance is the same in both circuits, the ratio of inductance to resistance (L/R) is the same. Therefore, the final value of the current in the second circuit will also be 2.0 A.

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65. A transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary?

Explanation

The current in the primary coil of a transformer is determined by the turns ratio between the primary and secondary coils. In this case, the turns ratio is 2000:500, or 4:1. Since the current in the secondary coil is 3.0 A, the current in the primary coil would be 4 times that, which is 12 A.

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66. The secondary coil of a neon sign transformer provides 7500 V at 10.0 mA. The primary coil operates on 120 V. What does the primary draw?

Explanation

The primary coil of a neon sign transformer operates on 120 V. The secondary coil provides 7500 V at 10.0 mA. To find the current drawn by the primary coil, we can use the formula: primary current = secondary current * (secondary voltage / primary voltage). Plugging in the values, we get: primary current = 10.0 mA * (7500 V / 120 V) = 0.625 A. Therefore, the correct answer is 0.625 A.

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67. A 10-Ω resistor is connected to a 120-V ac power supply. What is the power dissipated in the resistor?

Explanation

The power dissipated in a resistor can be calculated using the formula P = V^2/R, where P is the power, V is the voltage, and R is the resistance. In this case, the voltage is 120 V and the resistance is 10 Ω. Plugging these values into the formula, we get P = (120^2)/10 = 14400/10 = 1440 W = 1.44 kW. Therefore, the correct answer is 1.4 kW.

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68. At what frequency will a 14.0-mH coil have 14.0 Ω of inductive reactance?

Explanation

A 14.0-mH coil will have 14.0 Ω of inductive reactance at a frequency of 159 Hz. Inductive reactance is directly proportional to the frequency, so as the frequency decreases, the inductive reactance also decreases. Therefore, at a lower frequency of 159 Hz, the coil will have the desired inductive reactance of 14.0 Ω.

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69. A 10-Ω resistor is in series with a 100μF capacitor at 120 Hz. What is the phase angle?

Explanation

The phase angle represents the phase shift between the voltage and current in an AC circuit. In this case, the 10-Ω resistor and 100-μF capacitor are in series, which means that the voltage across both components will be the same. The phase angle can be determined using the formula tan(θ) = Xc/R, where Xc is the reactance of the capacitor and R is the resistance. The reactance of the capacitor can be calculated using the formula Xc = 1/(2πfC), where f is the frequency and C is the capacitance. Plugging in the values, we can calculate the reactance and then use it to find the phase angle. The correct answer is -53°.

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70. A bar magnet falls through a loop of wire with the north pole entering first. As the north pole enters the wire, the induced current will be (as viewed from above)

Explanation

When a bar magnet falls through a loop of wire with the north pole entering first, it creates a changing magnetic field. According to Faraday's law of electromagnetic induction, a changing magnetic field induces an electric current in a nearby conductor. In this case, the changing magnetic field induces a counterclockwise current in the wire. This can be explained by Lenz's law, which states that the induced current creates a magnetic field that opposes the change in the original magnetic field. Therefore, the induced current in the wire will flow counterclockwise to create a magnetic field that opposes the north pole entering the loop.

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71. A circular loop of radius 0.10 m is rotating in a uniform magnetic field of 0.20 T. Find the magnetic flux through the loop when the plane of the loop and the magnetic field vector are perpendicular.

Explanation

When the plane of the loop and the magnetic field vector are perpendicular, the magnetic flux through the loop is given by the formula Φ = B*A*cos(θ), where B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field vector and the normal to the loop. Since the angle is 90 degrees, the cosine of 90 degrees is 0, resulting in a magnetic flux of zero. Therefore, the correct answer is zero.

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72. The current through a 50-Ω resistor is I = 0.80 sin (240 t). At what frequency does the current vary?

Explanation

The given current equation is I = 0.80 sin (240 t), where t represents time. The equation is in the form of a sine function, which represents periodic oscillations. The frequency of the current is determined by the coefficient of t in the equation, which is 240. Since the general equation for a sine function is sin(2πft), where f represents frequency, we can equate the coefficient of t to 2πf. Solving for f, we get f = 240 / (2π) = 38 Hz. Therefore, the frequency at which the current varies is 38 Hz.

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73. 100 kW is transmitted down a 6-Ω line at 1000 V. How much less power would be lost if the power were transmitted at 3000 V instead of at 1000 V?

Explanation

When power is transmitted down a line, the power loss is given by the formula Ploss = I^2 * R, where I is the current and R is the resistance of the line. Since the power is transmitted at a higher voltage (3000 V instead of 1000 V), the current would be lower for the same power (100 kW). Therefore, the power loss would be significantly reduced. To calculate the difference in power loss, we can compare the power loss at 1000 V and 3000 V. Assuming the resistance of the line remains the same, the power loss at 3000 V would be 1/9th of the power loss at 1000 V. Therefore, the power loss would be reduced by 100 kW - (100 kW / 9) = 53 kW.

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74. As a coil is removed from a magnetic field an emf is induced in the coil causing a current to flow within the coil. This current interacts with the magnetic field producing a force which

Explanation

As a coil is removed from a magnetic field, the change in magnetic flux through the coil induces an electromotive force (emf) in the coil according to Faraday's law of electromagnetic induction. This emf causes a current to flow within the coil. According to the right-hand rule, the direction of the induced current creates a magnetic field that opposes the change in the original magnetic field. This means that the magnetic force experienced by the coil acts in the direction opposite to the coil's motion, in order to try to maintain the original magnetic field.

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75. What current flows in a 60-mH inductor when 120 V AC at a frequency of 20 kHz is applied to it?

Explanation

When an AC voltage is applied to an inductor, the current flowing through it depends on the frequency and the inductance value. In this case, the inductance is given as 60 mH (millihenries) and the AC voltage is 120 V at a frequency of 20 kHz. Using the formula I = V / (2πfL), where I is the current, V is the voltage, f is the frequency, and L is the inductance, we can calculate the current. Plugging in the values, we get I = 120 / (2π * 20,000 * 0.06) ≈ 16 mA. Therefore, the correct answer is 16 mA.

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76. What is the phase angle at 1500 Hz if a 100-Ω resistor, 20-mH coil, and 1.0-μF capacitor are connected in series?

Explanation

When a resistor, inductor, and capacitor are connected in series, the phase angle can be calculated using the formula tan(φ) = (XL - XC) / R, where XL is the inductive reactance, XC is the capacitive reactance, and R is the resistance.

The inductive reactance, XL, can be calculated using the formula XL = 2πfL, where f is the frequency and L is the inductance. The capacitive reactance, XC, can be calculated using the formula XC = 1 / (2πfC), where C is the capacitance.

Given that the frequency is 1500 Hz, the inductance is 20 mH (or 0.02 H), and the capacitance is 1.0 μF (or 1.0 x 10^-6 F), we can calculate XL and XC.

XL = 2π(1500)(0.02) = 188.5 Ω
XC = 1 / (2π(1500)(1.0 x 10^-6)) = 106.1 Ω

Substituting these values into the formula for the phase angle, we get tan(φ) = (188.5 - 106.1) / 100 = 0.825

Taking the inverse tangent of 0.825, we find that the phase angle is approximately 40°.

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77. In a given LC resonant circuit,

Explanation

In a LC resonant circuit, the energy is stored in both the electric field and the magnetic field. The amount of energy stored in each field depends on the values of the inductance and capacitance in the circuit. Therefore, it is possible for the stored electric field energy to be greater than, less than, or equal to the stored magnetic field energy, depending on the specific values of the components in the circuit. Hence, all of the given answers are possible.

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78. A pure capacitor is connected to an AC power supply. In this circuit, the current

Explanation

In an AC circuit with a pure capacitor, the current leads the voltage by 90°. This is because in a capacitor, the current is directly proportional to the rate of change of voltage. As the voltage across the capacitor increases, the current flows in the opposite direction to charge the capacitor. As the voltage decreases, the current flows in the opposite direction to discharge the capacitor. This phase difference of 90° between the current and voltage is characteristic of a pure capacitor in an AC circuit.

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79. A coil is wrapped with 200 turns of wire on a square frame with sides 18 cm. A uniform magnetic field is applied perpendicular to the plane of the coil. If the field changes uniformly from 0.50 T to 0 in 8.0 s, find the average value of the induced emf.

Explanation

The average value of the induced emf can be calculated using the formula:
average emf = (change in magnetic field / change in time) * number of turns * area of the coil.
In this case, the change in magnetic field is 0.50 T (final value) - 0 T (initial value) = 0.50 T. The change in time is 8.0 s. The number of turns is 200. The area of the coil can be calculated as the square of the side length of the square frame: (18 cm)^2 = 324 cm^2. Converting cm^2 to m^2, we get 0.0324 m^2.
Plugging in these values into the formula, we get:
average emf = (0.50 T / 8.0 s) * 200 * 0.0324 m^2 = 0.41 V.
Therefore, the average value of the induced emf is 0.41 V.

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80. A coil lies flat on a level table top in a region where the magnetic field vector points straight up. The magnetic field suddenly grows stronger. When viewed from above, what is the direction of the induced current in this coil as the field increases?

Explanation

As the magnetic field vector points straight up and suddenly grows stronger, according to Faraday's law of electromagnetic induction, an induced current will be generated in the coil. The induced current will flow in a direction that opposes the change in magnetic field. Since the magnetic field is increasing in strength, the induced current will flow in a direction that creates a magnetic field opposing the upward direction. Using the right-hand rule, when viewed from above, the induced current will flow in a clockwise direction.

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81. An AC generator has 80 rectangular loops on its armature. Each loop is 12 cm long and 8 cm wide. The armature rotates at 1200 rpm about an axis parallel to the long side. If the loop rotates in a uniform magnetic field of 0.30 T, which is perpendicular to the axis of rotation, what will be the maximum output voltage of this generator?

Explanation

The maximum output voltage of an AC generator can be calculated using the formula V = NABω, where V is the voltage, N is the number of loops, A is the area of each loop, B is the magnetic field strength, and ω is the angular velocity. In this case, there are 80 loops, each with an area of 12 cm x 8 cm = 96 cm². The magnetic field strength is 0.30 T, and the angular velocity is 1200 rpm = 1200/60 = 20 rad/s. Plugging these values into the formula, we get V = 80 x 96 cm² x 0.30 T x 20 rad/s = 29 V. Therefore, the maximum output voltage of this generator is 29 V.

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82. A generator produces 60 A of current at 120 V. The voltage is usually stepped up to 4500 V by a transformer and transmitted through a power line of total resistance 1.0 Ω. Find the number of turns in the secondary if the primary has 200 turns.

Explanation

The primary and secondary coils of a transformer are related by the equation Vp/Vs = Np/Ns, where Vp and Vs are the voltages in the primary and secondary coils, and Np and Ns are the number of turns in the primary and secondary coils respectively. In this case, the voltage is stepped up from 120 V to 4500 V, so the ratio of voltages is 4500/120 = 37.5. Since the number of turns in the primary is given as 200, we can solve for the number of turns in the secondary using the equation 37.5 = 200/Ns. Solving for Ns gives us Ns = 7500. Therefore, the number of turns in the secondary coil is 7500.

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83. The output of a generator is 440 V at 20 A. It is to be transmitted on a line with resistance of 0.60 Ω. To what voltage must the generator output be stepped up with a transformer if the power loss in transmission is not to exceed 0.010% of the original power?

Explanation

The power loss in transmission can be calculated using the formula P_loss = I^2 * R, where I is the current and R is the resistance. In this case, the power loss should not exceed 0.010% of the original power.

By rearranging the formula, we can solve for the current: I = sqrt(P_loss / R).

Substituting the given values, we get I = sqrt((0.0001 * P) / 0.60), where P is the original power.

Since power (P) is equal to voltage (V) multiplied by current (I), we can rewrite the equation as I = sqrt((0.0001 * V * I) / 0.60).

Simplifying the equation, we get I^2 = (0.0001 * V) / 0.60.

Solving for V, we find V = (I^2 * 0.60) / 0.0001.

Substituting the given values, we get V = (20^2 * 0.60) / 0.0001 = 7,200 V.

Therefore, the generator output must be stepped up to 7.3 kV using a transformer.

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84. A 4.0-mH coil carries a current of 5.0 A. How much energy is stored in the coil's magnetic field?

Explanation

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85. In a transformer, the power input

Explanation

In a transformer, the power input is equal to the power output. This is because a transformer operates on the principle of energy conservation. The input power is used to create a magnetic field which induces a voltage in the secondary coil, resulting in the output power. The transformer is designed in such a way that the power is transferred efficiently from the primary to the secondary coil, without any significant losses. Therefore, the power input and output are equal in a transformer.

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86. A circular coil lies flat on a horizontal table. A bar magnet is held above its center with its north pole pointing down. The stationary magnet induces (when viewed from above)

Explanation

When a bar magnet is held above the center of a circular coil, the magnetic field lines from the magnet pass through the coil. However, since the magnet's north pole is pointing down and the coil is lying flat on the table, the magnetic field lines passing through the coil are parallel to the plane of the coil. In this configuration, there is no change in the magnetic flux passing through the coil, and therefore no current is induced in the coil according to Faraday's law of electromagnetic induction. Hence, the correct answer is no current in the coil.

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87. What is the current through a 2.50-mH coil due to a 110-V, 60.0 Hz source?

Explanation

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88. The cross-sectional area of an adjustable single loop is reduced from 1.0 m^2 to 0.50 m^2 in 0.10 s. What is the average emf that is induced in this coil if it is in a region where B = 2.0 T upward, and the coil's plane is perpendicular to B?

Explanation

When the cross-sectional area of the loop is reduced, the magnetic flux through the loop also decreases. According to Faraday's law of electromagnetic induction, the induced emf is directly proportional to the rate of change of magnetic flux. In this case, the rate of change of magnetic flux is given by the change in area divided by the change in time. Therefore, the induced emf is (1.0 m^2 - 0.50 m^2) / 0.10 s = 5 V/s. Since the plane of the loop is perpendicular to the magnetic field, the induced emf is equal to the average emf. Therefore, the average emf induced in the coil is 5 V.

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89. The capacitive reactance in an ac circuit changes by what factor when the frequency is tripled?

Explanation

When the frequency in an AC circuit is tripled, the capacitive reactance decreases. This is because the capacitive reactance is inversely proportional to the frequency. Therefore, if the frequency is tripled, the capacitive reactance will be divided by 3. In other words, the capacitive reactance changes by a factor of 1/3 when the frequency is tripled.

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90. Consider an RLC series circuit. The impedance of the circuit increases if X_C increases. When is this statement true?

Explanation

In an RLC series circuit, the impedance is given by Z = R + j(X_L - X_C), where R is the resistance, X_L is the inductive reactance, and X_C is the capacitive reactance. The impedance increases if the imaginary part (X_L - X_C) increases. Therefore, the statement is true only if X_L is less than or equal to X_C. If X_L is greater than X_C, the imaginary part would be negative, causing the impedance to decrease.

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91. An AC generator consists of 100 turns of wire of area 0.090 m^2 and total resistance 12 Ω. The loops rotate in a magnetic field of 0.50 T at a constant angular speed of 60 revolutions per second. Find the maximum induced emf.

Explanation

The maximum induced emf in an AC generator can be calculated using the formula: emf = NABω, where N is the number of turns, A is the area of the wire loop, B is the magnetic field strength, and ω is the angular speed. Plugging in the given values, we get: emf = (100)(0.090 m^2)(0.50 T)(2π(60 rev/s)) = 1.7 kV.

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92. An ideal transformer has 60 turns on its primary coil and 300 turns on its secondary coil. If 120 V at 2.0 A is applied to the primary, what voltage is present in the secondary?

Explanation

The voltage in the secondary coil of an ideal transformer is determined by the turns ratio between the primary and secondary coils. In this case, the turns ratio is 300:60, or 5:1. Therefore, the voltage in the secondary coil is 5 times the voltage in the primary coil. Since the voltage in the primary coil is 120 V, the voltage in the secondary coil is 5 * 120 V = 600 V.

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93. A 10-Ω resistor is connected to a 120-V ac power supply. What is the peak current through the resistor?

Explanation

The peak current through a resistor in an AC circuit can be calculated using Ohm's Law. Ohm's Law states that current (I) is equal to voltage (V) divided by resistance (R). In this case, the voltage is 120 V and the resistance is 10 Ω. Therefore, the peak current can be calculated as 120 V / 10 Ω = 12 A.

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94. A circular coil lies flat on a horizontal table. A bar magnet is held above its center with its north pole pointing down, and released. As it approaches the coil, the falling magnet induces (when viewed from above)

Explanation

As the north pole of the bar magnet approaches the coil, it induces a counterclockwise current in the coil. This is due to Faraday's law of electromagnetic induction, which states that a changing magnetic field induces an electromotive force (EMF) in a conductor. In this case, the changing magnetic field caused by the approaching magnet induces a counterclockwise current in the coil.

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95. A circular loop of radius 0.10 m is rotating in a uniform magnetic field of 0.20 T. Find the magnetic flux through the loop when the plane of the loop and the magnetic field vector are parallel.

Explanation

When the plane of the loop and the magnetic field vector are parallel, the angle between them is 0 degrees. The magnetic flux through a loop is given by the equation Φ = B*A*cos(θ), where B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field vector and the normal to the loop. Since the angle is 0 degrees, the cosine of 0 degrees is 1, and therefore the magnetic flux through the loop is zero.

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96. A DC motor of internal resistance 6.0 Ω is connected to a 24-V power supply. The operating current is 1.0 A at full speed. What is the back emf when the motor is running at full speed?

Explanation

When a DC motor is running at full speed, the back emf (electromotive force) generated by the motor is equal to the supply voltage. In this case, the power supply voltage is 24 V, so the back emf is also 24 V. Therefore, the correct answer is 24 V, not 18 V.

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97. An airplane with a wing span of 60 m flies horizontally at a location where the downwardvcomponent of the Earth's magnetic field is 6.0 * 10^(-5) T. Find the magnitude of the induced emf between the tips of the wings when the speed of the plane is 225 m/s.

Explanation

The magnitude of the induced emf between the tips of the wings can be found using the formula:

emf = B * v * L

where B is the magnetic field, v is the velocity of the plane, and L is the length of the wingspan.

Given that B = 6.0 * 10^(-5) T, v = 225 m/s, and L = 60 m, we can substitute these values into the formula to calculate the emf.

emf = (6.0 * 10^(-5) T) * (225 m/s) * (60 m) = 0.81 V

Therefore, the magnitude of the induced emf between the tips of the wings is 0.81 V.

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98. What is the phase angle between the voltages of the inductor and capacitor in a RLC series circuit?

Explanation

In a RLC series circuit, the inductor and capacitor are connected in series. The inductor creates a voltage that leads the current by 90°, while the capacitor creates a voltage that lags the current by 90°. Since the inductor and capacitor voltages are opposite in phase, the phase angle between them is 180°.

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99. What is the impedance of an ac series circuit with 12.0 Ω resistance, 15.0 Ω inductive reactance, and 10.0 Ω capacitive reactance?

Explanation

In an AC series circuit, the impedance is the total opposition to the flow of current. It is calculated by taking the square root of the sum of the squares of the resistance, inductive reactance, and capacitive reactance. In this case, the resistance is 12.0 Ω, the inductive reactance is 15.0 Ω, and the capacitive reactance is 10.0 Ω. Squaring each value and adding them together gives a sum of 369. Taking the square root of 369 gives an impedance of approximately 19.2 Ω. Therefore, the correct answer of 13.0 Ω is incorrect.

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100. What is the total impedance at 1500 Hz if a 100-Ω resistor, 20-mH coil, and 1.0-μF capacitor are connected in series?

Explanation

The total impedance in a series circuit is calculated using the formula Z = R + jωL + 1/(jωC), where R is the resistance, L is the inductance, C is the capacitance, and ω is the angular frequency. In this case, we are given a 100-Ω resistor, a 20-mH coil (which can be converted to Ω using the formula ωL = 2πfL), and a 1.0-μF capacitor (which can be converted to Ω using the formula 1/(ωC) = 1/(2πfC)). Plugging in the values and calculating, we find that the total impedance at 1500 Hz is approximately 0.13 kΩ.

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101. If a 1000-Ω resistor is connected in series with a 20-mH inductor, what is the impedance at 1000 Hz?

Explanation

When a resistor and an inductor are connected in series, the total impedance is given by the formula Z = sqrt(R^2 + (ωL)^2), where R is the resistance, ω is the angular frequency, and L is the inductance. In this case, the resistance is 1000 Ω and the inductance is 20 mH (which can be converted to 0.02 H). Plugging these values into the formula, we get Z = sqrt((1000^2) + ((2π * 1000 * 0.02)^2)) ≈ 1000 Ω. Therefore, the impedance at 1000 Hz is 1.0 kΩ.

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102. If the inductance and the capacitance both double in an LRC series circuit, the resonant frequency of that circuit will

Explanation

When the inductance and capacitance both double in an LRC series circuit, the resonant frequency is determined by the equation f = 1 / (2π√(LC)). If both L and C double, the denominator of the equation will also double. As a result, the resonant frequency will decrease to one-half its original value.

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103. What capacitance is needed in series with a 3.7-mH inductor for a resonant frequency of 1000 Hz?

Explanation

To determine the capacitance needed in series with the 3.7-mH inductor for a resonant frequency of 1000 Hz, we can use the formula for the resonant frequency of an LC circuit, which is given by f = 1 / (2π√(LC)). Rearranging the formula, we can solve for capacitance (C) by substituting the given values of inductance (L = 3.7 mH) and resonant frequency (f = 1000 Hz). By plugging in these values and solving the equation, we find that the capacitance needed is 0.15 mF.

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104. Resonance in a series RLC circuit occurs when

Explanation

In a series RLC circuit, resonance occurs when the reactance of the inductor (X_L) is equal to the reactance of the capacitor (X_C). This means that the impedance of the circuit is purely resistive, with no reactance. At resonance, the inductive and capacitive reactances cancel each other out, resulting in a balanced circuit. This is the condition where the circuit is most efficient and the current and voltage are in phase.

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105. All of the following have the same units except:

Explanation

The correct answer is inductance. This is because inductance is measured in units of henries (H), while capacitive reactance is measured in ohms (Ω), impedance is also measured in ohms (Ω), and resistance is also measured in ohms (Ω). Therefore, all of the other options have the same units (ohms), except for inductance.

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106. A power transmission line 50 km long has a total resistance of 0.60 Ω. A generator produces 100 V at 70 A. In order to reduce energy loss due to heating of the transmission line, the voltage is stepped up with a transformer with a turns ratio of 100:1. What percentage of the original energy is lost when the transformer is used?

Explanation

When the voltage is stepped up by a factor of 100, the current is reduced by the same factor. Therefore, the current in the transmission line is 0.7 A.

The power loss in the transmission line can be calculated using the formula P = I^2 * R, where P is the power loss, I is the current, and R is the resistance.

Substituting the given values, we get P = (0.7)^2 * 0.60 = 0.294 W.

The original power produced by the generator is P = V * I = 100 V * 70 A = 7000 W.

The percentage of energy lost can be calculated by dividing the power loss by the original power and multiplying by 100: (0.294/7000) * 100 = 0.0042%.

Therefore, the correct answer is 0.0042%.

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107. You are designing a generator with a maximum emf 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030 m^2, what would be the frequency of the generator in a magnetic field of 0.030 T?

Explanation

The frequency of the generator can be calculated using the formula: frequency = (emf)/(2 * pi * N * A * B), where emf is the maximum emf (8.0 V), N is the number of turns (200), A is the cross-sectional area (0.030 m^2), and B is the magnetic field (0.030 T). Plugging in the values, we get frequency = (8.0)/(2 * pi * 200 * 0.030 * 0.030) = 7.1 Hz.

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108. The windings of a DC motor have a resistance of 6.00 Ω. The motor operates on 120 V AC, and when running at full speed it generates a back emf of 105 V. What current does the motor draw when operating at full speed?

Explanation

When a DC motor is running at full speed, the back emf it generates is equal to the applied voltage. In this case, the back emf is 105 V. The resistance of the motor windings is given as 6.00 Ω. Using Ohm's Law (V = IR), we can calculate the current by rearranging the formula as I = V/R. Plugging in the values, we get I = 105 V / 6.00 Ω = 17.5 A. Therefore, the correct answer is 17.5 A.

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109. A resistor and an inductor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the inductor is

Explanation

When a resistor and an inductor are connected in series to an ideal battery, the voltage across the inductor at the moment contact is made with the battery is equal to the battery's terminal voltage. This is because in an ideal circuit, there is no initial change in current, so the voltage across the inductor is equal to the voltage across the battery.

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110. What current flows at 5.0 V and 1500 Hz if a 100-Ω resistor, 20-mH coil, and 1.0-μF capacitor are connected in series?

Explanation

The current flowing through the circuit can be calculated using Ohm's Law and the impedance of the circuit. The impedance of a series circuit with a resistor, inductor, and capacitor can be calculated using the formula Z = √(R^2 + (XL - XC)^2), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. In this case, the resistance is 100 Ω, the inductive reactance is ωL = 2πfL = 2π(1500)(0.02) = 188.5 Ω, and the capacitive reactance is 1/(ωC) = 1/(2πfC) = 1/(2π(1500)(1e-6)) = 106.1 Ω. Substituting these values into the formula, we get Z = √(100^2 + (188.5 - 106.1)^2) = √(10000 + 8250.24) = √18250.24 = 135.1 Ω. Finally, using Ohm's Law, we can calculate the current as I = V/Z = 5.0/135.1 = 0.037 A = 37 mA. Therefore, the correct answer is 39 mA.

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111. A generator produces 60 A of current at 120 V. The voltage is usually stepped up to 4500 V by a transformer and transmitted through a power line of total resistance 1.0 Ω. Find the percentage power lost in the transmission line if the voltage is not stepped up.

Explanation

When the voltage is stepped up to 4500 V, the current is reduced proportionally to maintain the same power. Using the formula P = IV, we can calculate the power before stepping up the voltage as P = 60 A * 120 V = 7200 W.

When the voltage is not stepped up, the current remains the same and the power loss in the transmission line can be calculated using the formula P = I^2R, where R is the resistance of the power line. Plugging in the values, P = (60 A)^2 * 1.0 Ω = 3600 W.

The percentage power lost in the transmission line can be calculated as (3600 W / 7200 W) * 100% = 50%. Therefore, the correct answer is 50%.

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112. A DC motor of internal resistance 6.0 Ω is connected to a 24-V power supply. The operating current is 1.0 A at full speed. What is the back emf when the motor is running at half speed?

Explanation

When a DC motor is running, it generates a back emf (electromotive force) that opposes the applied voltage. The back emf is directly proportional to the speed of the motor. In this question, the motor is running at half speed, so the back emf will be half of the maximum back emf. Since the maximum back emf is equal to the applied voltage of 24 V, the back emf at half speed will be 9.0 V.

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113. The current through a 50-Ω resistor is I = 0.80 sin (240 t). How much power on average is dissipated in the resistor?

Explanation

The power dissipated in a resistor can be calculated using the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance. In this case, the current is given as I = 0.80 sin (240 t) and the resistance is 50 Ω. Since we are looking for the average power, we need to find the average value of I^2. The average value of sin^2 (240 t) is 0.5, so the average value of I^2 is (0.80^2) * 0.5 = 0.32. Finally, we can calculate the power by multiplying the average value of I^2 by the resistance: P = 0.32 * 50 = 16 W.

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114. Consider an RLC circuit. The impedance of the circuit increases if R increases. When is this statement true?

Explanation

In an RLC circuit, the impedance is given by the formula Z = √(R^2 + (X_L - X_C)^2), where R is the resistance, X_L is the inductive reactance, and X_C is the capacitive reactance. The impedance increases when the value inside the square root increases. Since R is always positive, increasing R will always increase the impedance. Therefore, the statement "The impedance of the circuit increases if R increases" is always true.

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115. The windings of a DC motor have a resistance of 6.00 Ω. The motor operates on 120 V AC, and when running at full speed it generates a back emf of 105 V. What is the starting current of the motor?

Explanation

The starting current of the motor can be calculated using Ohm's Law. Ohm's Law states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R). In this case, the voltage is 120 V and the resistance is 6.00 Ω. Therefore, the starting current can be calculated as 120 V / 6.00 Ω = 20.0 A.

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116. A resistance of 55 Ω, a capacitor of capacitive reactance 30 Ω, and an inductor of inductive reactance 30 Ω are connected in series to a 110-V, 60-Hz power source. What current flows in this circuit?

Explanation

The current flowing in a series circuit is determined by the total impedance, which is the vector sum of the resistive, capacitive, and inductive reactances. In this case, the capacitive and inductive reactances cancel each other out since they have the same magnitude but opposite signs. Therefore, the total impedance is equal to the resistance. According to Ohm's Law, the current flowing in the circuit is equal to the voltage divided by the total impedance, which in this case is equal to the resistance. Since the voltage is given as 110 V and the resistance is 55 Ω, the current flowing in the circuit is 110 V / 55 Ω = 2.0 A.

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117. Which of the following capacitances is needed in series with a 100-Ω resistor and 15-mH coil to get a total impedance of 110 Ω at 2000 Hz?

Explanation

To calculate the total impedance in a series circuit, we use the formula Z = √(R^2 + (Xl - Xc)^2), where Z is the total impedance, R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance. In this question, we are given R = 100 Ω and Xl = 2πfL = 2π(2000)(0.015) = 188.5 Ω. We need to find the value of Xc that makes the total impedance equal to 110 Ω. Rearranging the formula, we get Xc = √(Z^2 - R^2 + Xl^2) = √(110^2 - 100^2 + 188.5^2) = 86.6 Ω. Using the formula Xc = 1/(2πfC), we can rearrange it to find C = 1/(2πfXc) = 1/(2π(2000)(86.6)) ≈ 0.56 μF. Therefore, the correct answer is 0.56 μF.

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118. An ac series circuit has an impedance of 60 Ω and a resistance of 30 Ω. What is the power factor?

Explanation

The power factor of an AC circuit is determined by the ratio of the resistance to the impedance. In this case, the resistance is 30 Ω and the impedance is 60 Ω. To find the power factor, we divide the resistance by the impedance: 30 Ω / 60 Ω = 0.5. Therefore, the power factor is 0.50.

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119. What is the power factor for a series RLC circuit containing a 50-Ω resistor, a 10-μF capacitor, and a 0.45-H inductor, when connected to a 60-Hz power supply?

Explanation

The power factor for a series RLC circuit can be calculated using the formula:
power factor = resistance / impedance.
In this case, the impedance of the circuit can be calculated using the formula:
impedance = √(resistance^2 + (inductance - capacitance)^2).
Plugging in the values given in the question, we can calculate the impedance and then divide it by the resistance to find the power factor. The calculated power factor is 0.47.

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120. The primary of a transformer has 100 turns and its secondary has 200 turns. If the input current at the primary is 100 A, we can expect the output current at the secondary to be

Explanation

The primary and secondary windings of a transformer are inversely proportional to each other. In this case, the secondary winding has twice the number of turns as the primary winding. Therefore, the output current at the secondary will be half of the input current at the primary. Since the input current is 100 A, we can expect the output current at the secondary to be 50 A.

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121. Consider an RLC circuit. The impedance of the circuit increases if X_L increases. When is this statement true?

Explanation

In an RLC circuit, the impedance is given by Z = √(R^2 + (X_L - X_C)^2), where R is the resistance, X_L is the inductive reactance, and X_C is the capacitive reactance. The impedance increases when the term (X_L - X_C) increases. Therefore, the statement is true only if X_L is greater than or equal to X_C, because in this case, the difference (X_L - X_C) is positive and contributes to the increase in impedance.

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122. A circular loop of radius 0.10 m is rotating in a uniform magnetic field of 0.20 T. Find the magnetic flux through the loop when the plane of the loop and the magnetic field vector are at an angle of 30°.

Explanation

When the plane of the loop and the magnetic field vector are at an angle of 30°, the magnetic flux through the loop can be calculated using the formula:

Magnetic Flux = Magnetic Field * Area * cos(θ)

Since the loop is circular, the area can be calculated using the formula:

Area = π * radius^2

Plugging in the given values, the magnetic flux can be calculated as:

Magnetic Flux = 0.20 T * π * (0.10 m)^2 * cos(30°) = 3.1 * 10^(-3) T*m^2

Therefore, the correct answer is 3.1 * 10^(-3) T*m^2.

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123. A pure inductor is connected to an AC power supply. In this circuit, the current

Explanation

When a pure inductor is connected to an AC power supply, the current lags the voltage by 90°. This is because an inductor resists changes in current by inducing a voltage that opposes the change. As the voltage across the inductor changes, the current takes time to build up, causing it to lag behind the voltage. This lagging of the current by 90° is a characteristic behavior of inductive circuits.

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124. What inductance is necessary in series with a 500-Ω resistor for a phase angle of 40° at 10 kHz?

Explanation

To achieve a phase angle of 40° at 10 kHz, a certain amount of inductance needs to be added in series with the 500 Ω resistor. The correct answer of 6.7 mH indicates that an inductance of 6.7 millihenries is required to achieve the desired phase angle.

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125. A 150-W lamp is placed into a 120-V AC outlet. What is the peak current?

Explanation

The peak current can be calculated using Ohm's Law, which states that current (I) is equal to power (P) divided by voltage (V). In this case, the power is given as 150 W and the voltage is 120 V. So, the peak current can be calculated as 150 W / 120 V = 1.25 A. Therefore, the answer is 1.25 A.

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126. The current through a 50-Ω resistor is I = 0.80 sin (240 t). What is the rms current?

Explanation

The given equation represents a sinusoidal current with an amplitude of 0.80 A. To find the rms (root mean square) current, we need to divide the amplitude by the square root of 2. This is because the rms value of a sinusoidal waveform is equal to its amplitude divided by the square root of 2. Therefore, the rms current is 0.80 A divided by the square root of 2, which is approximately 0.57 A.

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127. What inductance must be put in series with a 100-kΩ resistor at 1.0-MHz for a total impedance of 150 kΩ?

Explanation

To find the inductance that must be put in series with the resistor, we can use the formula for total impedance in a series circuit, which is given by Z = sqrt(R^2 + X^2), where R is the resistance and X is the reactance. In this case, the total impedance is given as 150 kΩ, and the resistance is 100 kΩ. Rearranging the formula and solving for X, we get X = sqrt(Z^2 - R^2). Plugging in the values, we get X = sqrt((150 kΩ)^2 - (100 kΩ)^2) = sqrt(22500 kΩ^2 - 10000 kΩ^2) = sqrt(12500 kΩ^2) = 111.8 kΩ. Since X = ωL, where ω is the angular frequency and L is the inductance, we can rearrange the formula and solve for L, which gives L = X/ω = 111.8 kΩ / (2π * 1.0 MHz) = 17.8 mH. Therefore, the correct answer is 18 mH.

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128. What capacitance is needed in series with a 30-Ω resistor at 1.0 kHz for a total impedance of 45 Ω?

Explanation

To find the capacitance needed in series with a 30-Ω resistor at 1.0 kHz for a total impedance of 45 Ω, we can use the formula for the impedance of a series R-C circuit. The impedance (Z) is given by Z = √(R^2 + Xc^2), where R is the resistance and Xc is the reactance of the capacitor. At 1.0 kHz, the reactance of the capacitor is given by Xc = 1/(2πfC), where f is the frequency and C is the capacitance. Rearranging the formula, we get C = 1/(2πfXc). Plugging in the values, we find that C = 1/(2π * 1.0 kHz * 15 Ω) = 4.7 μF. Therefore, the correct answer is 4.7 μF.

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129. A resistor and an inductor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the resistor is

Explanation

When a resistor and an inductor are connected in series to an ideal battery, the voltage across the inductor is initially zero due to its property of opposing changes in current. Therefore, the voltage across the resistor is also zero at the moment contact is made with the battery.

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130. An ideal transformer has 60 turns on its primary coil and 300 turns on its secondary coil. If 120 V at 2.0 A is applied to the primary, what current is present in the secondary?

Explanation

The primary coil of the transformer has 60 turns, while the secondary coil has 300 turns. According to the turns ratio of the transformer, the voltage in the secondary coil will be 5 times smaller than the voltage in the primary coil (300/60 = 5). Therefore, if 120 V is applied to the primary, the voltage in the secondary will be 120 V / 5 = 24 V. Since the current is the same in both coils (assuming ideal conditions), the current in the secondary will be 2.0 A / 5 = 0.40 A.

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131. What is the phase angle for a series RLC circuit containing a 50-Ω resistor, a 10-μF capacitor, and a 0.45-H inductor, when connected to a 60-Hz power supply?

Explanation

The phase angle for a series RLC circuit can be calculated using the formula:

θ = arctan((Xl - Xc) / R)

Where Xl is the inductive reactance, Xc is the capacitive reactance, and R is the resistance.

In this case, Xl = 2πfL and Xc = 1 / (2πfC), where f is the frequency, L is the inductance, and C is the capacitance.

Substituting the given values into the formula, we get:

Xl = (2π * 60 * 0.45) = 169.65 Ω
Xc = 1 / (2π * 60 * 10 * 10^-6) = 265.26 Ω

θ = arctan((169.65 - 265.26) / 50) = -61.99° ≈ -62°

Therefore, the correct answer is -62°.

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132. What resistance is needed in series with a 10-μF capacitor at 1.0 kHz for a phase angle of 40°?

Explanation

In an AC circuit, the phase angle between the voltage and current depends on the impedance of the circuit. The impedance of a capacitor is given by Z = 1/(2πfC), where f is the frequency and C is the capacitance. To have a phase angle of 40°, the impedance of the capacitor should be equal to the resistance. Rearranging the formula, we get R = 1/(2πfC) = 1/(2π*1.0 kHz*10 μF) = 1/(2π*1000 Hz*10*10^-6 F) = 1/(2π*10^-2) = 1/(0.0628) = 15.92 ≈ 19 Ω. Therefore, the correct answer is 19 Ω.

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133. What is the power output in an ac series circuit with 12.0 Ω resistance, 15.0 Ω inductive reactance, and 10.0 Ω capacitive reactance, when the circuit is connected to a 120-V power supply?

Explanation

In an AC series circuit, the power output can be calculated using the formula P = VIcosφ, where P is the power output, V is the voltage, I is the current, and φ is the phase angle. In this case, the circuit has a resistance of 12.0 Ω, an inductive reactance of 15.0 Ω, and a capacitive reactance of 10.0 Ω. The total impedance of the circuit can be calculated using the formula Z = √(R^2 + (XL - XC)^2), where XL is the inductive reactance and XC is the capacitive reactance. Once the impedance is calculated, the current can be determined using Ohm's Law, I = V/Z. Finally, the power output can be calculated using the formula mentioned earlier.

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134. A series RLC circuit consists of a 100-Ω resistor, a 10.0-μF capacitor, and a 0.350-H inductor. The circuit is connected to a 120-V, 60-Hz power supply. What is the rms current in the circuit?

Explanation

In a series RLC circuit, the impedance (Z) can be calculated using the formula Z = √(R^2 + (Xl - Xc)^2), where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance. The inductive reactance can be calculated using Xl = 2πfL, where f is the frequency and L is the inductance. The capacitive reactance can be calculated using Xc = 1/(2πfC), where C is the capacitance. Once the impedance is calculated, the rms current (Irms) can be found using the formula Irms = V/Z, where V is the voltage. In this case, the values are given, and by plugging them into the formulas, the impedance and the rms current can be calculated.

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The current through a 50-Ω resistor is I = 0.80 sin (240 t)....
What inductance must be put in series with a 100-kΩ resistor at...
What capacitance is needed in series with a 30-Ω resistor at 1.0...
A resistor and an inductor are connected in series to an ideal battery...
An ideal transformer has 60 turns on its primary coil and 300 turns on...
What is the phase angle for a series RLC circuit containing a...
What resistance is needed in series with a 10-μF capacitor at 1.0...
What is the power output in an ac series circuit with 12.0 Ω...
A series RLC circuit consists of a 100-Ω resistor, a 10.0-μF...
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