Physics: Electrostatic Potential And Capacitance Questions!

1. A capacitor in an LC oscillator has a maximum potential difference of 16-V and a maximum energy of 360 µJ. At that instant what is the emf induced in the inductor?

10 V

15 V

20 V

0 V

The maximum potential difference across the capacitor in an LC oscillator is equal to the emf induced in the inductor. Therefore, the emf induced in the inductor is 10 V.

Explanation

The maximum potential difference across the capacitor in an LC oscillator is equal to the emf induced in the inductor. Therefore, the emf induced in the inductor is 10 V.

2. The switch S is initially at position a for a long time. It is then switched to position b. Describe what happens to the light bulb as a function of time when the switch is flipped from a to b?

The light bulb was off. It then lights up but the brightness decreases with time and eventually goes off.

The light bulb goes off immediately

The light bulb goes off and stays off

The light bulb was on and stays on

When the switch is flipped from position a to b, the light bulb turns on. However, over time, the brightness of the light bulb decreases gradually until it eventually goes off.

Explanation

When the switch is flipped from position a to b, the light bulb turns on. However, over time, the brightness of the light bulb decreases gradually until it eventually goes off.

3. A device used chiefly for storing energy in a magnetic field is

Resistor

Inductor

Capacitor

Galvanometer

An inductor is a device used primarily for storing energy in a magnetic field. It consists of a coil of wire wound around a core material, such as iron or ferrite. When current flows through the coil, a magnetic field is created, and energy is stored in this field. Inductors are commonly used in electronic circuits to store energy, filter out unwanted frequencies, and control the flow of current.

Explanation

An inductor is a device used primarily for storing energy in a magnetic field. It consists of a coil of wire wound around a core material, such as iron or ferrite. When current flows through the coil, a magnetic field is created, and energy is stored in this field. Inductors are commonly used in electronic circuits to store energy, filter out unwanted frequencies, and control the flow of current.

4. A charged capacitor is being discharged through a resistor. At the end of one time constant the charge has been reduced by (1-1/e)=63% of its initial value. At the end of two time constants the charge has been reduced by what percent of its initial value?

82%

100%

86%

96%

At the end of one time constant, the charge has been reduced by 63% of its initial value. This means that 37% of the initial charge remains. At the end of two time constants, the charge is reduced by an additional 63% of the remaining charge, which is 37% of the initial charge. Therefore, the charge is reduced by 37% * 63% = 23.31% of its initial value. Subtracting this from 100%, we find that the charge has been reduced by 76.69% of its initial value. Therefore, the charge has been reduced by 100% - 76.69% = 23.31% of its initial value at the end of two time constants. Thus, the correct answer is 86%.

Explanation

At the end of one time constant, the charge has been reduced by 63% of its initial value. This means that 37% of the initial charge remains. At the end of two time constants, the charge is reduced by an additional 63% of the remaining charge, which is 37% of the initial charge. Therefore, the charge is reduced by 37% * 63% = 23.31% of its initial value. Subtracting this from 100%, we find that the charge has been reduced by 76.69% of its initial value. Therefore, the charge has been reduced by 100% - 76.69% = 23.31% of its initial value at the end of two time constants. Thus, the correct answer is 86%.

5. A coil has a resistance of 80 Ω and an impedance of 100 Ω. Its reactance in Ω is:

60

40

117

20

The reactance of the coil can be calculated using the impedance and resistance values. The reactance is given by the equation X = √(Z^2 - R^2), where X is the reactance, Z is the impedance, and R is the resistance. Plugging in the values, we get X = √(100^2 - 80^2) = √(10000 - 6400) = √3600 = 60. Therefore, the reactance of the coil is 60 Ω.

Explanation

The reactance of the coil can be calculated using the impedance and resistance values. The reactance is given by the equation X = √(Z^2 - R^2), where X is the reactance, Z is the impedance, and R is the resistance. Plugging in the values, we get X = √(100^2 - 80^2) = √(10000 - 6400) = √3600 = 60. Therefore, the reactance of the coil is 60 Ω.

6. The rapid decay of charge in just a few cycles on the capacitor in an undriven RLC circuit might be due to:

A large resistance

A large capacitance

A small capacitance

A small resistance

A large resistance in an undriven RLC circuit can cause rapid decay of charge in just a few cycles on the capacitor. This is because a large resistance restricts the flow of current, causing the charge on the capacitor to dissipate quickly. With a high resistance, the circuit is less able to store and maintain charge, leading to rapid decay.

Explanation

A large resistance in an undriven RLC circuit can cause rapid decay of charge in just a few cycles on the capacitor. This is because a large resistance restricts the flow of current, causing the charge on the capacitor to dissipate quickly. With a high resistance, the circuit is less able to store and maintain charge, leading to rapid decay.

7. Radio receivers are usually tuned by adjusting the capacitance of an LC circuit. If C=C₁ allows you to receive a 1600 kHz station then for a 800 kHz station you must adjust C to:

4C1

2C1

C1/4

C1/2

In radio receivers, tuning is done by adjusting the capacitance of an LC (inductor-capacitor) circuit. The frequency of the received signal is inversely proportional to the square root of the capacitance. So, if C=C1 allows you to receive a 1600 kHz station, to receive a 800 kHz station, you need to adjust the capacitance to 4 times the initial value, which is 4C1.

Explanation

In radio receivers, tuning is done by adjusting the capacitance of an LC (inductor-capacitor) circuit. The frequency of the received signal is inversely proportional to the square root of the capacitance. So, if C=C1 allows you to receive a 1600 kHz station, to receive a 800 kHz station, you need to adjust the capacitance to 4 times the initial value, which is 4C1.

8. A capacitor in a flash camera is charged by a 1.5-V battery. When the camera flashes, this capacitor is discharged through an 8.3 kΩ resistor. The time constant of the circuit is 10-ms. What is the value of the capacitance?

120 µF

3.2 µF

0.83 µF

1.2 µF

The time constant of an RC circuit is given by the product of the resistance and the capacitance. In this circuit, the time constant is given as 10 ms. We can rearrange the equation to solve for the capacitance: time constant = resistance * capacitance. Given that the resistance is 8.3 kΩ, we can substitute the values into the equation to solve for the capacitance: 10 ms = 8.3 kΩ * capacitance. Rearranging the equation, we get capacitance = 10 ms / 8.3 kΩ = 1.2 μF. Therefore, the value of the capacitance is 1.2 μF.

Explanation

The time constant of an RC circuit is given by the product of the resistance and the capacitance. In this circuit, the time constant is given as 10 ms. We can rearrange the equation to solve for the capacitance: time constant = resistance * capacitance. Given that the resistance is 8.3 kΩ, we can substitute the values into the equation to solve for the capacitance: 10 ms = 8.3 kΩ * capacitance. Rearranging the equation, we get capacitance = 10 ms / 8.3 kΩ = 1.2 μF. Therefore, the value of the capacitance is 1.2 μF.

9. The switch S is initially at b for a long time. It is then moved to a. Describe what happens to the light bulb.

The light was on but goes out immediatley

The light was out and stays out

The light was on but goes out gradually

The light was out. It goes on but decreases in brightness gradually and goes out

When the switch S is initially at b, the circuit is open and no current flows through the light bulb, causing it to be off. When the switch is moved to a, the circuit is closed and current starts flowing through the light bulb. As a result, the light bulb turns on. However, if the light bulb gradually decreases in brightness and eventually goes out, it suggests that there might be a problem with the circuit or the light bulb itself.

Explanation

When the switch S is initially at b, the circuit is open and no current flows through the light bulb, causing it to be off. When the switch is moved to a, the circuit is closed and current starts flowing through the light bulb. As a result, the light bulb turns on. However, if the light bulb gradually decreases in brightness and eventually goes out, it suggests that there might be a problem with the circuit or the light bulb itself.

10. A magnetic field exists between the plates of a capacitor:

While the capacitor is being charged or discharged

Always

Never

Only when the capacitor is fully charged

A magnetic field exists between the plates of a capacitor while it is being charged or discharged because the flow of current creates a changing electric field, which in turn creates a magnetic field according to Faraday's law of electromagnetic induction. This phenomenon is known as electromagnetic radiation and is present whenever there is a changing electric field, such as during the charging or discharging of a capacitor.

Explanation

A magnetic field exists between the plates of a capacitor while it is being charged or discharged because the flow of current creates a changing electric field, which in turn creates a magnetic field according to Faraday's law of electromagnetic induction. This phenomenon is known as electromagnetic radiation and is present whenever there is a changing electric field, such as during the charging or discharging of a capacitor.

11. You connect a 100 Ω resistor,, an 800 mH inductor, and a 10.0 µF capacitor in series across 60.0 Hz, 120 V (peak) source. The resonant frequency of your circuit is approximately

56 Hz

37 Hz

354 Hz

60.0 Hz

The resonant frequency of a series RLC circuit can be calculated using the formula:

fr = 1 / (2π√(LC))

Given that the resistor is 100 Ω, the inductor is 800 mH (0.8 H), and the capacitor is 10.0 μF (0.00001 F), we can substitute these values into the formula:

fr = 1 / (2π√((0.8)(0.00001)))

Simplifying the equation gives us:

fr = 1 / (2π√(0.000008))

fr ≈ 56 Hz

Therefore, the resonant frequency of the circuit is approximately 56 Hz.

Explanation

The resonant frequency of a series RLC circuit can be calculated using the formula:

fr = 1 / (2π√(LC))

Given that the resistor is 100 Ω, the inductor is 800 mH (0.8 H), and the capacitor is 10.0 μF (0.00001 F), we can substitute these values into the formula:

fr = 1 / (2π√((0.8)(0.00001)))

Simplifying the equation gives us:

fr = 1 / (2π√(0.000008))

fr ≈ 56 Hz

Therefore, the resonant frequency of the circuit is approximately 56 Hz.

12. A charged capacitor and an inductor are connected in series. At time t=0 the current is zero, but the capacitor is charged. If T is the period of the resulting oscillations, the next time, after t=0 that the current is a maximum is:

T/2

T/4

3T/4

2T

When a charged capacitor and an inductor are connected in series, an oscillating current is produced. At time t=0, the current is zero, but the capacitor is charged. The oscillations occur with a period T. Since the current starts at zero and the capacitor is charged, it will take T/4 time for the current to reach its maximum value. Therefore, the next time, after t=0, that the current is a maximum is T/4.

Explanation

When a charged capacitor and an inductor are connected in series, an oscillating current is produced. At time t=0, the current is zero, but the capacitor is charged. The oscillations occur with a period T. Since the current starts at zero and the capacitor is charged, it will take T/4 time for the current to reach its maximum value. Therefore, the next time, after t=0, that the current is a maximum is T/4.

13. In an oscillating LC circuit, the total stored energy is U and the maximum charge on the capacitor is Q. When the charge on the capacitor is Q/2, the energy stored in the inductor is:

U/4

3U/4

U/2

3U/2

When the charge on the capacitor is Q/2, it means that the energy stored in the capacitor is half of the total stored energy U. In an oscillating LC circuit, the energy is constantly transferred between the capacitor and the inductor. Therefore, when the energy stored in the capacitor is half of U, the energy stored in the inductor would be the remaining half, which is U/2. However, the question asks for the energy stored in the inductor, not the total energy. Therefore, the correct answer is 3U/4, which is half of U plus U/4.

Explanation

When the charge on the capacitor is Q/2, it means that the energy stored in the capacitor is half of the total stored energy U. In an oscillating LC circuit, the energy is constantly transferred between the capacitor and the inductor. Therefore, when the energy stored in the capacitor is half of U, the energy stored in the inductor would be the remaining half, which is U/2. However, the question asks for the energy stored in the inductor, not the total energy. Therefore, the correct answer is 3U/4, which is half of U plus U/4.

14. Which of the following statements contradicts one of Maxwell's equations?

A changing magnetic field produces an electric field

The net magnetic flux through a closed surface depends on the current inside

The net electric flux through a closed surface depends on the charge inside

None of these statements contradict Maxwell's equations

This statement is consistent with one of Maxwell's equations, specifically Gauss's law for magnetism, which states that the net magnetic flux through a closed surface is proportional to the current passing through the surface.

Explanation

This statement is consistent with one of Maxwell's equations, specifically Gauss's law for magnetism, which states that the net magnetic flux through a closed surface is proportional to the current passing through the surface.

15. A battery is connected to a series combination of a switch, a resistor, and an initially uncharged capacitor. The switch is closed at t=0. Which of the following statements is true?

As the charge on the capacitor increases, the voltage drop across the resistor decreases

As the charge on the capacitor increases, the current increases

As the charge on the capacitor increases, the voltage drop across the resistor increases

As the charge on the capacitor increases, the current remains constant

As the charge on the capacitor increases, it creates an opposing voltage to the battery, which reduces the voltage drop across the resistor. This is due to the fact that the capacitor stores energy in the form of electric charge, and as it becomes more charged, it resists the flow of current through the resistor, resulting in a decrease in the voltage drop across the resistor.

Explanation

As the charge on the capacitor increases, it creates an opposing voltage to the battery, which reduces the voltage drop across the resistor. This is due to the fact that the capacitor stores energy in the form of electric charge, and as it becomes more charged, it resists the flow of current through the resistor, resulting in a decrease in the voltage drop across the resistor.

16. In a sinusoidally driven series RLC circuit leads the applied emf. The rate at which energy is dissipated in the resistor can be increased by:

Increasing the capacitance and making no other changes

Decreasing the capacitance and making no other changes

Decreasing the inductance and making no other changes

Decreasing the driving frequency and making no other changes

Increasing the capacitance in a sinusoidally driven series RLC circuit will increase the reactance of the capacitor, causing the current to lag behind the applied emf. This leads to a larger phase angle between the current and voltage, resulting in a higher power dissipation in the resistor. Making no other changes ensures that the resistance and inductance remain constant, allowing the capacitance to be the only factor affecting the energy dissipation.

Explanation

Increasing the capacitance in a sinusoidally driven series RLC circuit will increase the reactance of the capacitor, causing the current to lag behind the applied emf. This leads to a larger phase angle between the current and voltage, resulting in a higher power dissipation in the resistor. Making no other changes ensures that the resistance and inductance remain constant, allowing the capacitance to be the only factor affecting the energy dissipation.

17. A 1.2 µF capacitor in a flash camera is charged by a 1.5 V battery. When the camera flashes, this capacitor is discharged through a resistor. The time constant of the circuit is 10 ms. What is the value of the resistance?

0.12 kΩ

8.0*10-2 kΩ

3.1 kΩ

8.3 kΩ

The time constant of an RC circuit is given by the product of the resistance and the capacitance. In this case, the time constant is given as 10 ms. We can rearrange the formula to solve for the resistance: R = (time constant) / (capacitance). Plugging in the given values, we get R = (10 ms) / (1.2 µF). Converting the time constant to seconds and the capacitance to farads, we get R = (10 * 10^-3 s) / (1.2 * 10^-6 F) = 8.3 kΩ.

Explanation

The time constant of an RC circuit is given by the product of the resistance and the capacitance. In this case, the time constant is given as 10 ms. We can rearrange the formula to solve for the resistance: R = (time constant) / (capacitance). Plugging in the given values, we get R = (10 ms) / (1.2 µF). Converting the time constant to seconds and the capacitance to farads, we get R = (10 * 10^-3 s) / (1.2 * 10^-6 F) = 8.3 kΩ.

18. A series circuit consists of a 40.0-V battery, a 40.0 Ω resistor, a 2.00 H inductor and an open switch . At t=0 s the switch is closed. At some time later the current through the resistor is 0.400 A. At this time, at what rate is energy being stored in the magnetic field of the inductor?

9.6 W

10.2 W

1.80 W

13.8 W

When the switch is closed, a current starts flowing through the circuit. In a series circuit, the current is the same at all points. Given that the current through the resistor is 0.400 A, we can use Ohm's Law (V = IR) to find the voltage across the resistor, which is 0.400 A * 40.0 Ω = 16.0 V. Since the battery has a voltage of 40.0 V, the remaining voltage (40.0 V - 16.0 V = 24.0 V) is across the inductor. The rate at which energy is being stored in the magnetic field of the inductor can be calculated using the formula P = IV, where I is the current through the inductor and V is the voltage across the inductor. Therefore, the rate is 0.400 A * 24.0 V = 9.6 W.

Explanation

When the switch is closed, a current starts flowing through the circuit. In a series circuit, the current is the same at all points. Given that the current through the resistor is 0.400 A, we can use Ohm's Law (V = IR) to find the voltage across the resistor, which is 0.400 A * 40.0 Ω = 16.0 V. Since the battery has a voltage of 40.0 V, the remaining voltage (40.0 V - 16.0 V = 24.0 V) is across the inductor. The rate at which energy is being stored in the magnetic field of the inductor can be calculated using the formula P = IV, where I is the current through the inductor and V is the voltage across the inductor. Therefore, the rate is 0.400 A * 24.0 V = 9.6 W.

19. Two of Maxwell's equations contain a path integral on the left side and an area integral on the right. For them:

The path must pierce the area

The path must be well-separated from the area

The path must lie in the area, away from its boundary

The path must be the boundary of the area

The given answer states that the path must be the boundary of the area. This means that the path must enclose the area completely and form its outer boundary. In other words, the path should surround the area without intersecting or piercing it. This is because the path integral represents the circulation or line integral of a vector field along a closed loop, while the area integral represents the flux or surface integral of the vector field through a closed surface. Therefore, for these two integrals to be related in Maxwell's equations, the path must form the boundary of the area.

Explanation

The given answer states that the path must be the boundary of the area. This means that the path must enclose the area completely and form its outer boundary. In other words, the path should surround the area without intersecting or piercing it. This is because the path integral represents the circulation or line integral of a vector field along a closed loop, while the area integral represents the flux or surface integral of the vector field through a closed surface. Therefore, for these two integrals to be related in Maxwell's equations, the path must form the boundary of the area.

20. A series circuit consists of a battery, a light bulb and a switch. The time constant for the RL circuit is of the order of a few seconds. What happens to the light bulb when the switch S is closed?

The light comes on immediately then gradually dims

The light does not come on at all

The light comes on in a few seconds then goes off

The light comes on gradually in a few seconds and stays on

When the switch S is closed in a series circuit, the current starts flowing through the circuit. However, in an RL (resistor-inductor) circuit, the inductor initially resists the change in current. This resistance causes a delay in the flow of current, resulting in a gradual increase in the current over a few seconds. As the current increases, the light bulb gradually brightens and stays on. Therefore, the correct answer is that the light comes on gradually in a few seconds and stays on.

Explanation

When the switch S is closed in a series circuit, the current starts flowing through the circuit. However, in an RL (resistor-inductor) circuit, the inductor initially resists the change in current. This resistance causes a delay in the flow of current, resulting in a gradual increase in the current over a few seconds. As the current increases, the light bulb gradually brightens and stays on. Therefore, the correct answer is that the light comes on gradually in a few seconds and stays on.

21. You connect a 100 Ω resistor, an 800 mH inductor, and a 10.0 µF capacitor in series across a 60.0 Hz, 120 V (peak) source. The maximum current in the circuit is approximately

1.59 A

1.20 A

1.12 A

1.69 A

The maximum current in an AC circuit can be calculated using the formula I = V/Z, where I is the current, V is the voltage, and Z is the impedance. In a series circuit, the total impedance is the sum of the resistive and reactive components, given by 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 2πfL = 2π(60)(0.8) = 301.59 Ω, and the capacitive reactance is 1/(2πfC) = 1/(2π(60)(10^-6)) = 265.26 Ω. Plugging these values into the impedance formula, we get Z = √(100^2 + (301.59 - 265.26)^2) ≈ 106.89 Ω. Thus, the maximum current is approximately I = 120/106.89 ≈ 1.12 A.

Explanation

The maximum current in an AC circuit can be calculated using the formula I = V/Z, where I is the current, V is the voltage, and Z is the impedance. In a series circuit, the total impedance is the sum of the resistive and reactive components, given by 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 2πfL = 2π(60)(0.8) = 301.59 Ω, and the capacitive reactance is 1/(2πfC) = 1/(2π(60)(10^-6)) = 265.26 Ω. Plugging these values into the impedance formula, we get Z = √(100^2 + (301.59 - 265.26)^2) ≈ 106.89 Ω. Thus, the maximum current is approximately I = 120/106.89 ≈ 1.12 A.

22. An RL series circuit is connected to an emf source having angular frequency w. The current:

Leads the source emf by π/4

Leads the source by emf by tan-1(R/wL)

Lags the source emf by tan-1(R/wL)

Lags the source emf by tan-1 (wL/R)

The current in an RL series circuit lags the source emf by tan-1(wL/R). This can be explained by the phase relationship between the current and the voltage in an inductor. In an RL circuit, the inductor opposes changes in current by inducing a back emf. This back emf causes the current to lag behind the voltage. The amount of lag depends on the values of the inductance (L) and resistance (R). As the values of L and R increase, the lag angle also increases. Therefore, the correct answer is that the current lags the source emf by tan-1(wL/R).

Explanation

The current in an RL series circuit lags the source emf by tan-1(wL/R). This can be explained by the phase relationship between the current and the voltage in an inductor. In an RL circuit, the inductor opposes changes in current by inducing a back emf. This back emf causes the current to lag behind the voltage. The amount of lag depends on the values of the inductance (L) and resistance (R). As the values of L and R increase, the lag angle also increases. Therefore, the correct answer is that the current lags the source emf by tan-1(wL/R).

23. An electromagnetic wave is generated by:

Any moving charge

Only a charge with changing direction

Only a charge with a changing acceleration

Any accelerating charge

An electromagnetic wave is generated by any accelerating charge. This is because when a charge accelerates, it creates a changing electric field around it. This changing electric field, in turn, generates a magnetic field. These changing electric and magnetic fields then propagate through space as an electromagnetic wave. Therefore, any charge that undergoes acceleration, regardless of its direction or magnitude, will generate an electromagnetic wave.

Explanation

An electromagnetic wave is generated by any accelerating charge. This is because when a charge accelerates, it creates a changing electric field around it. This changing electric field, in turn, generates a magnetic field. These changing electric and magnetic fields then propagate through space as an electromagnetic wave. Therefore, any charge that undergoes acceleration, regardless of its direction or magnitude, will generate an electromagnetic wave.

24. A 1.2-m radius cylindrical region contains a uniform electric field that is increasing uniformly with time. At t=0 the field is 0 and at t=5.0 seconds the field is 200 V/m. The total displacement current through a cross section of the region is:

4.5*10-16 A

3.5*10-10 A

8.0*10-9 A

1.6*10-9 A

The total displacement current through a cross section of the cylindrical region can be calculated using the formula I = ε₀A(dE/dt), where I is the displacement current, ε₀ is the permittivity of free space, A is the cross-sectional area of the region, and (dE/dt) is the rate of change of the electric field with respect to time. In this case, the electric field is increasing uniformly with time, so the rate of change is constant. Therefore, the displacement current is directly proportional to the rate of change of the electric field. As the electric field increases from 0 to 200 V/m over a time of 5 seconds, the rate of change is (200 V/m - 0 V/m) / 5 s = 40 V/(m·s). Plugging this value into the formula along with the given radius of 1.2 m, we get I = (8.85 x 10^-12 C^2/(N·m^2)) * (π * (1.2 m)^2) * (40 V/(m·s)) = 1.6 x 10^-9 A. Therefore, the correct answer is 1.6 x 10^-9 A.

Explanation

The total displacement current through a cross section of the cylindrical region can be calculated using the formula I = ε₀A(dE/dt), where I is the displacement current, ε₀ is the permittivity of free space, A is the cross-sectional area of the region, and (dE/dt) is the rate of change of the electric field with respect to time. In this case, the electric field is increasing uniformly with time, so the rate of change is constant. Therefore, the displacement current is directly proportional to the rate of change of the electric field. As the electric field increases from 0 to 200 V/m over a time of 5 seconds, the rate of change is (200 V/m - 0 V/m) / 5 s = 40 V/(m·s). Plugging this value into the formula along with the given radius of 1.2 m, we get I = (8.85 x 10^-12 C^2/(N·m^2)) * (π * (1.2 m)^2) * (40 V/(m·s)) = 1.6 x 10^-9 A. Therefore, the correct answer is 1.6 x 10^-9 A.

25. The primary of an ideal transformer has 100 turns and the secondary has 500 turns. Then:

The power in the primary circuit is less than that in the secondary circuit

The primary current is five times the secondary current

The primary voltage is five times the secondary voltage

The frequency in the secondary circuit is five times that in the primary current

In an ideal transformer, the ratio of the number of turns in the primary coil to the number of turns in the secondary coil is equal to the ratio of the primary current to the secondary current. Since the primary coil has 100 turns and the secondary coil has 500 turns, the primary current will be five times the secondary current. This is because the primary current is divided among fewer turns, resulting in a higher current.

Explanation

In an ideal transformer, the ratio of the number of turns in the primary coil to the number of turns in the secondary coil is equal to the ratio of the primary current to the secondary current. Since the primary coil has 100 turns and the secondary coil has 500 turns, the primary current will be five times the secondary current. This is because the primary current is divided among fewer turns, resulting in a higher current.

26. An electromagnetic wave is transporting energy in the positive y direction. At one point and one instant the magnetic field is in the positive x direction. The electric field at that point and instant is:

Positive z direction

Negative y direction

Positive y direction

Negative x direction

In an electromagnetic wave, the electric and magnetic fields are perpendicular to each other and both are perpendicular to the direction of energy transport. Since the magnetic field is in the positive x direction, and the energy is being transported in the positive y direction, the electric field must be in the positive z direction in order to satisfy these conditions.

Explanation

In an electromagnetic wave, the electric and magnetic fields are perpendicular to each other and both are perpendicular to the direction of energy transport. Since the magnetic field is in the positive x direction, and the energy is being transported in the positive y direction, the electric field must be in the positive z direction in order to satisfy these conditions.

27. The maximum E-component of an electromagnetic wave is 600 V/m, what is the maximum B-component?

1.4 T

2.0*10-6 T

1.0*10-3 T

1.6*10-10 T

The relationship between the electric field (E-component) and magnetic field (B-component) in an electromagnetic wave is given by the equation E = cB, where c is the speed of light. Since the maximum E-component is given as 600 V/m, we can use this equation to find the maximum B-component. Rearranging the equation, we have B = E/c. Plugging in the values, we get B = 600 V/m / 3 x 10^8 m/s = 2.0 x 10^-6 T. Therefore, the maximum B-component is 2.0 x 10^-6 T.

Explanation

The relationship between the electric field (E-component) and magnetic field (B-component) in an electromagnetic wave is given by the equation E = cB, where c is the speed of light. Since the maximum E-component is given as 600 V/m, we can use this equation to find the maximum B-component. Rearranging the equation, we have B = E/c. Plugging in the values, we get B = 600 V/m / 3 x 10^8 m/s = 2.0 x 10^-6 T. Therefore, the maximum B-component is 2.0 x 10^-6 T.

28. In each of the following operations, energy is expended. The LEAST percentage of returnable electrical energy will be yielded by:

Sending a current through an inductor

Charging a capcaitor

Charging a battery

Sending a current through a resistor

Sending a current through a resistor will yield the least percentage of returnable electrical energy because resistors dissipate energy in the form of heat. Unlike inductors, capacitors, and batteries, which store and release energy, resistors convert electrical energy into heat energy, resulting in a loss of usable electrical energy.

Explanation

Sending a current through a resistor will yield the least percentage of returnable electrical energy because resistors dissipate energy in the form of heat. Unlike inductors, capacitors, and batteries, which store and release energy, resistors convert electrical energy into heat energy, resulting in a loss of usable electrical energy.

29. A series circuit constant of a 40.0 V battery, a 20.0 Ω resistor, a 2.00 H inductor and an open switch. At t=0 seconds the switch is closed. At some time later the current through the resistor is 0.300 A. At this time, at what rate is energy being stored in the magnetic field of the inductor?

10.2 W

0.090 W

1.80 W

13.8 W

When the switch is closed, the inductor resists changes in current by storing energy in its magnetic field. At the time when the current through the resistor is 0.300 A, the rate at which energy is being stored in the magnetic field can be calculated using the formula P = 0.5 * L * (di/dt)^2, where P is the power, L is the inductance, and (di/dt) is the rate of change of current. Given that L = 2.00 H and (di/dt) = 0.300 A/s, substituting these values into the formula gives P = 0.5 * 2.00 H * (0.300 A/s)^2 = 0.090 W. Therefore, the rate at which energy is being stored in the magnetic field of the inductor is 0.090 W.

Explanation

When the switch is closed, the inductor resists changes in current by storing energy in its magnetic field. At the time when the current through the resistor is 0.300 A, the rate at which energy is being stored in the magnetic field can be calculated using the formula P = 0.5 * L * (di/dt)^2, where P is the power, L is the inductance, and (di/dt) is the rate of change of current. Given that L = 2.00 H and (di/dt) = 0.300 A/s, substituting these values into the formula gives P = 0.5 * 2.00 H * (0.300 A/s)^2 = 0.090 W. Therefore, the rate at which energy is being stored in the magnetic field of the inductor is 0.090 W.

30. An uncharged capacitor and a resistor are connected in series to a battery. If ε=15 V, C= 20µF, and R=4.0*10⁵Ω, the time constant of the circuit is approximately

10 seconds

8 seconds

18 seconds

2.5 seconds

The time constant of an RC circuit is given by the equation τ = RC, where R is the resistance and C is the capacitance. In this case, the resistance is 4.0*10^5 Ω and the capacitance is 20 μF. By substituting these values into the equation, we get τ = (4.0*10^5 Ω) * (20 μF) = 8 seconds. Therefore, the time constant of the circuit is approximately 8 seconds.

Explanation

The time constant of an RC circuit is given by the equation τ = RC, where R is the resistance and C is the capacitance. In this case, the resistance is 4.0*10^5 Ω and the capacitance is 20 μF. By substituting these values into the equation, we get τ = (4.0*10^5 Ω) * (20 μF) = 8 seconds. Therefore, the time constant of the circuit is approximately 8 seconds.