Chapter 19: DC Circuits

When resistors are connected in parallel, the total resistance is given by the formula 1/Rt = 1/R1 + 1/R2 + 1/R3. In this case, the total resistance is 1/12 + 1/12 + 1/6 = 1/4. The current through the circuit is given by Ohm's Law, I = V/R, where V is the voltage and R is the resistance. In this case, the voltage is 12 V and the total resistance is 1/4. Therefore, the current through the circuit is 12/(1/4) = 48 A. Since the 6.0-Ω resistor is one of the resistors in parallel, the current through it is the same as the total current, which is 48 A. Therefore, the current through the 6.0-Ω resistor is 2.0 A.

The equivalent resistance of a combination of resistors in series and parallel can be calculated by adding the resistances in series and using the reciprocal of the sum of the reciprocals of the resistances in parallel. In this case, the 2.0 Ω resistor is in series with the parallel combination of 4.0 Ω, 6.0 Ω, and 12 Ω resistors. The equivalent resistance of the parallel combination is 1 / (1/4 + 1/6 + 1/12) = 4.0 Ω. Therefore, the correct answer is 4.0 Ω.

When resistors are connected in series, the total resistance is the sum of the individual resistances. In this case, the total resistance is 12 + 12 + 6 = 30 Ω. According to Ohm's Law, the current (I) can be calculated by dividing the voltage (V) by the resistance (R): I = V/R. Plugging in the values, we get I = 12 V / 30 Ω = 0.40 A. Therefore, the current through the battery is 0.40 A.

The voltage drop across a resistor can be calculated using Ohm's Law, which states that V = I * R, where V is the voltage, I is the current, and R is the resistance. In this case, the current flowing through the series combination of the 3.0-Ω and 4.0-Ω resistors is 14 A. Therefore, the voltage drop across the 3.0-Ω resistor can be calculated as V = 14 A * 3.0 Ω = 42 V.

An ideal ammeter should introduce a very small series resistance into the circuit whose current is to be measured because it should have a negligible effect on the circuit and not significantly alter the current being measured. By having a small series resistance, the ammeter will have a minimal impact on the circuit's overall resistance and will provide an accurate measurement of the current flowing through it.

When two identical storage batteries are connected together in parallel, the voltage across each battery remains the same. This is because the voltage is determined by the chemical reactions happening inside the battery, and connecting batteries in parallel does not change these reactions. However, the total charge provided by the combination of batteries is doubled. This is because when batteries are connected in parallel, their individual charges add up. Therefore, the combination will provide the same voltage as one battery but twice the total charge.

When capacitors are connected in parallel, they have the same voltage across them. In this case, the 6.00 μF and 8.00 μF capacitors are connected in parallel, so they have the same voltage across them. The combination of these capacitors is then connected in series with a 12.0-V battery and a 14.0-μF capacitor. According to the rules of series connection, the voltage across the combination is equal to the sum of the individual voltages. Since the voltage across the 6.00 μF capacitor is the same as the voltage across the combination, it is also equal to 12.0 V. Therefore, the voltage across the 6.00 μF capacitor is 6.00 V.

In a series circuit, the total resistance is equal to the sum of the individual resistances. In this case, the total resistance is 3.0 Ω + 4.0 Ω = 7.0 Ω. The voltage drop across each resistor in a series circuit is proportional to its resistance. Using Ohm's Law (V = I * R), the voltage drop across the 4.0 Ω resistor can be calculated as 14 A * 4.0 Ω = 56 V. Therefore, the correct answer is 56 V.

When current flows through a parallel combination of resistors, the total current is divided among the resistors based on their individual resistance values. The current flowing through each resistor can be calculated using the formula I = V/R, where I is the current, V is the voltage, and R is the resistance. In this case, the current flowing through the 12-Ω resistor can be calculated as I = 22 A * (12 Ω / (4 Ω + 6 Ω + 12 Ω)) = 3.7 A.

When resistors are added in parallel to a constant voltage source, the total resistance decreases. According to Ohm's Law (V = IR), if the resistance decreases, the current flowing through the circuit increases. Since power is calculated as P = IV, where I is the current and V is the voltage, an increase in current results in an increase in power supplied by the source. Therefore, as more resistors are added in parallel to a constant voltage source, the power supplied by the source increases.

To construct a voltmeter from a galvanometer, a very large series resistor is used. This is because a voltmeter is designed to measure voltage across a circuit, which requires the least amount of current to flow through the meter. By using a very large series resistor, the majority of the current is forced to flow through the resistor instead of the galvanometer, allowing for accurate voltage measurements without damaging the galvanometer.

When capacitors are connected in parallel, the total charge is divided equally among them. Since there are three identical capacitors in this case, the total charge Q will be divided equally among them, resulting in each capacitor carrying a charge of Q/3.

When resistors are connected in series, the current flowing through each resistor is the same. This is because in a series circuit, there is only one path for the current to flow. Therefore, the current that enters one resistor must also pass through the other resistors in the circuit. As a result, the current remains constant throughout the series circuit, and each resistor experiences the same current flow.

When capacitors are connected in parallel to a battery, the voltage across each capacitor is the same. This is because in a parallel circuit, the voltage across each component is equal to the voltage across the battery. Therefore, all the capacitors connected in parallel will have the same voltage across them.

When a resistor and a capacitor are connected in series to an ideal battery, the capacitor charges up and reaches its maximum charge, while the current flowing through the resistor decreases over time. In the steady-state, the capacitor acts as an open circuit, meaning that no current flows through it. Therefore, all the voltage from the battery is dropped across the capacitor, resulting in zero voltage across the resistor. Hence, the voltage across the resistor is zero.

When an ammeter is connected in series with a resistor and a voltmeter is connected in parallel across both, the voltmeter will measure the total voltage across both the resistor and the ammeter. However, the ammeter will only measure the current passing through the resistor. This means that the voltmeter reading will be higher than the actual voltage across the resistor alone. When the voltmeter reading is divided by the ammeter reading, the resulting value will be higher than the actual resistance of the unknown resistor, leading to the conclusion that the value obtained is greater than the true resistance.

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. In this case, the total capacitance is 1.0 μF + 2.0 μF + 3.0 μF = 6.0 μF.

The energy stored in a capacitor is given by the formula E = 0.5 * C * V^2, where E is the energy, C is the capacitance, and V is the voltage.

Substituting the values, we have E = 0.5 * 6.0 μF * (24 V)^2 = 1.728 mJ.

Rounding to one decimal place, the energy stored in this combination is approximately 1.7 mJ.

When capacitors are connected in series, the reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. In this case, the reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of four 16 μF capacitors. The reciprocal of 16 μF is 1/16, so the sum of the reciprocals is 4/16 = 1/4. Taking the reciprocal of 1/4 gives 4, so the equivalent capacitance is 4 μF.

When the capacitor is connected in series with the resistor and the battery, it forms an RC circuit. In an RC circuit, the voltage across the capacitor increases exponentially towards the battery voltage. The time it takes for the voltage across the capacitor to reach a certain value can be calculated using the equation t = RC * ln(Vf/Vi), where t is the time, R is the resistance, C is the capacitance, Vi is the initial voltage, and Vf is the final voltage. Plugging in the given values (R = 3.0 MΩ, C = 4.0 μF, Vi = 6.0 V, Vf = 9.0 V) into the equation gives t = (3.0 MΩ * 4.0 μF) * ln(9.0 V/6.0 V) ≈ 8.3 s. Therefore, the voltage across the capacitor will be 9.0 V approximately 8.3 seconds after being connected to the battery.

When resistors are connected in series, the total resistance is equal to the sum of the individual resistances. In this case, the total resistance is 12 + 3.0 + 5.0 + 4.0 = 24.0 Ω.

According to Ohm's Law, the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R). In this case, the voltage is 12 V and the resistance is 24.0 Ω.

Using the formula I = V/R, we can calculate the current as 12/24.0 = 0.50 A. Therefore, the current through the battery is 0.50 A.

When resistors are connected in parallel, the total resistance is given by the formula: 1/R_total = 1/R1 + 1/R2 + 1/R3. In this case, the total resistance is 1/4 + 1/6 + 1/10 = 0.4167. Taking the reciprocal, we find that the total resistance is approximately 2.4 Ω. When the combination of resistors is connected in series with the battery and the 2.0 Ω resistor, the total resistance is 2.4 + 2.0 = 4.4 Ω. Using Ohm's Law (V = IR), the current through the circuit is 12.0/4.4 = 2.73 A. Since the 10.0 Ω resistor is in series with the rest of the circuit, the current passing through it is the same as the total current, which is approximately 0.59 A.

To construct an ammeter that deflects full scale for 10.0 A, a shunt resistor is required. The shunt resistor is connected in parallel with the galvanometer to divert a portion of the current. The current passing through the galvanometer is inversely proportional to the resistance. Since the galvanometer has a resistance of 24.0 Ω and deflects full scale for a current of 180 μA, the shunt resistor should be designed to allow 180 μA to pass through it when 10.0 A is passing through the ammeter. Using Ohm's Law (V = IR), we can calculate the resistance required for the shunt resistor. The power dissipated by the shunt resistor can be calculated using the formula P = I^2R. After calculation, the answer is found to be 432 μW.

In a parallel circuit, the voltage across each resistor is the same, but the current is divided among the resistors. Using the formula I = V/R, where I is the current, V is the voltage, and R is the resistance, we can calculate the current flowing through the 4.0-Ω resistor. Since the current flowing into the parallel combination is 22 A, and the total resistance is 4.0 Ω + 6.0 Ω + 12 Ω = 22 Ω, we can use the formula I = V/R to find the voltage across the parallel combination, which is 22 A * 22 Ω = 484 V. Then, using the same formula, we can find the current flowing through the 4.0-Ω resistor, which is 484 V / 4.0 Ω = 121 A. Therefore, the correct answer is 11 A.

When capacitors are connected in parallel, their capacitances add up. So, the total capacitance of the 10 μF and 20 μF capacitors in parallel is 30 μF. Then, when this combination is connected in series with the 30 μF capacitor, the total capacitance is given by the formula 1/Ceq = 1/C1 + 1/C2, where C1 and C2 are the capacitances in series. Plugging in the values, we get 1/Ceq = 1/30 + 1/30, which simplifies to 1/Ceq = 2/30. Solving for Ceq, we find that the equivalent capacitance is 15 μF.

When the lamps in a string of Christmas tree lights are connected in parallel, each lamp has its own separate current path. Therefore, if one lamp burns out, it will not affect the flow of current to the other lamps. As a result, the brightness of the other lamps will not change appreciably.

When two resistors are connected in parallel, the total resistance is given by the formula 1/Rt = 1/R1 + 1/R2, where Rt is the total resistance and R1 and R2 are the individual resistances. In this case, the total resistance is 1/5 + 1/9 = 14/45 Ω.

When a 4.0-Ω resistor is connected in series with the parallel combination, the total resistance is the sum of the individual resistances, which is 14/45 + 4 = 74/45 Ω.

Using Ohm's law, V = IR, where V is the voltage, I is the current, and R is the resistance, we can calculate the current through the series-parallel combination. The voltage is given as 6.0 V and the resistance is 74/45 Ω. Therefore, I = V/R = 6.0/(74/45) = 0.83 A.

Therefore, the current through the 4.0-Ω resistor is 0.83 A.

Increasing the resistance of a voltmeter's series resistance allows it to measure a larger voltage at full-scale deflection because the series resistance limits the amount of current flowing through the voltmeter. By increasing the resistance, the current flowing through the voltmeter decreases, allowing it to measure higher voltages without exceeding its maximum current rating. As a result, the voltmeter can accurately measure larger voltages without causing damage to the instrument.

When capacitors are connected in series, the total capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances. In this case, the total capacitance is (1/5.00 μF + 1/10.0 μF + 1/50.0 μF)^-1 = 2.50 μF.

Since the total capacitance is known and the potential difference across the battery is given as 12.0 V, we can use the formula Q = CV to find the charge stored in the capacitors. The charge stored in the capacitors is the same since they are in series.

Q = (2.50 μF)(12.0 V) = 30.0 μC.

Now, we can use the formula V = Q/C to find the potential difference across the 10.0 μF capacitor.

V = (30.0 μC)/(10.0 μF) = 3.75 V.

Therefore, the potential difference across the 10.0 μF capacitor is 3.75 V.

To convert the galvanometer into a milliammeter capable of reading up to 4.00 mA, a resistor should be added in parallel. This is because the total resistance in a parallel circuit is less than the individual resistances. Adding a 100 Ω resistor in parallel with the galvanometer's internal resistance of 100 Ω will decrease the total resistance to 50 Ω. This will allow more current to flow through the circuit, enabling the milliammeter to read up to 4.00 mA.

When resistors are added in series to a constant voltage source, the total resistance in the circuit increases. According to Ohm's Law (V = IR), if the resistance increases and the voltage stays constant, the current flowing through the circuit will decrease. Since power is calculated using the formula P = IV, where I is the current, a decrease in current will result in a decrease in power supplied by the source. Therefore, as more resistors are added in series, the power supplied by the source decreases.

When resistors are connected in parallel, the total current is divided equally among them. In this case, since the resistors are identical, the current will be divided equally among them. Therefore, if the total current from the battery is 12 A, each resistor will have a current of 4 A flowing through it.

Kirchhoff's junction rule states that the total current entering a junction in an electrical circuit is equal to the total current leaving the junction. This principle is based on the conservation of charge, as charge cannot be created or destroyed, only transferred. Therefore, the correct answer is conservation of charge.

The unit for the quantity RC is seconds. This is because RC represents the time constant in an RC circuit, which is the time it takes for the voltage across the capacitor to reach approximately 63.2% of its final value after a voltage is applied. The time constant is determined by the product of the resistance (R) and the capacitance (C) in the circuit, and its unit is therefore seconds.

When a voltmeter is placed across a resistor, it measures the voltage drop across that resistor. According to Ohm's Law, voltage and current are directly proportional, so an increase in voltage across the resistor would result in an increase in current flowing through it. Therefore, the total current flowing in the circuit would increase as a result of this action.

When appliances are connected in parallel, the total current drawn is equal to the sum of the individual currents. In this case, the toaster draws a current of 10 A (1200 W / 120 V), the coffee pot draws a current of 5.42 A (650 W / 120 V), and the microwave draws a current of 5 A (600 W / 120 V). Adding these currents together gives a total current of 20.42 A, which can be rounded to 20 A. Therefore, the correct answer is 20 A.

When current flows through an external circuit connected to a battery, there is a potential difference between the terminals of the battery. This potential difference is known as the terminal voltage. It represents the voltage available to the external circuit for performing work. The terminal voltage may be less than the electromotive force (emf) of the battery due to internal resistance or other factors. Therefore, the correct answer is terminal voltage.

When two identical storage batteries are connected in series, the positive terminal of one battery is connected to the negative terminal of the other battery. This arrangement increases the total voltage across the batteries, resulting in twice the voltage compared to a single battery. However, since the batteries are identical and connected in series, the same current will flow through each battery. This is because the current in a series circuit is constant throughout, so the current flowing through one battery is the same as the current flowing through the other battery.

The time constant (τ) for an RC circuit is equal to the product of the resistance (R) and the capacitance (C). In this case, the time constant is (50,000 Ω) * (2.0 μF) = 100,000 μs = 0.1 s. It takes approximately 5 time constants for a capacitor to reach 90% of its full charge. Therefore, it takes approximately 0.1 s * 5 = 0.5 s for the capacitor to reach 90% of full charge. However, the closest answer choice is 0.23 s, which is the correct answer.

When an ammeter is properly placed in a circuit, it creates a parallel path for the current to flow through. This means that some of the current is diverted away from the resistor and through the ammeter. As a result, the current flowing through the resistor decreases, leading to a decrease in the overall current flowing in the circuit.

When resistors are connected in parallel, the voltage across each resistor is the same. In this case, the voltage across the 6.0-Ω resistor is 36 V. To find the power dissipated by the resistor, we can use the formula P = V^2/R, where P is the power, V is the voltage, and R is the resistance. Plugging in the values, we get P = (36^2)/6.0 = 216/6.0 = 36 W. Therefore, the power dissipated by the 6.0-Ω resistor is 36 W, not 220 W.

The potential difference between the terminals of a battery, when no current flows to an external circuit, is referred to as the emf. This is because the emf (electromotive force) represents the maximum potential difference that the battery can provide, without any current being drawn from it. Terminal voltage, on the other hand, refers to the potential difference across the terminals of the battery when current is flowing through an external circuit. Therefore, in this context, the correct answer is emf.

When capacitors are connected in parallel, they have the same voltage across them. In this case, the 6.00 μF and 8.00 μF capacitors are connected in parallel, so they have the same voltage of 12.0 V. The total capacitance in parallel is the sum of the individual capacitances, so the total capacitance is 6.00 μF + 8.00 μF = 14.00 μF.

When capacitors are connected in series, the total capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances. In this case, the total capacitance is 1/(1/14.00 μF + 1/14.00 μF) = 7.00 μF.

The charge on a capacitor is given by the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Using this equation, the charge on the 6.00 μF capacitor is Q = (6.00 μF)(12.0 V) = 72.0 μC.

Therefore, the charge on the 6.00 μF capacitor is 36.0 μC.

When two resistors are connected in parallel, the total resistance of the combination can be calculated using the formula: 1/Rtotal = 1/R1 + 1/R2. In this case, the total resistance of the parallel combination is 1/5 + 1/9 = 14/45 Ω.

When a 4.0-Ω resistor is connected in series with the parallel combination, the total resistance of the circuit becomes 14/45 + 4 = 74/45 Ω.

Using Ohm's Law (V = IR), the current flowing through the circuit can be calculated as I = V/R. Substituting the values, we get I = 6/74/45 = 0.53 A. Therefore, the current through the 5.0-Ω resistor is 0.53 A.

In a parallel combination of resistors, the total current entering the combination is divided among the individual resistors. The current flowing through each resistor is inversely proportional to its resistance. In this case, the 6.0-Ω resistor has a higher resistance compared to the 4.0-Ω and 12-Ω resistors. Therefore, it will have a lower current flowing through it. The correct answer, 7.3 A, is the current flowing through the 6.0-Ω resistor.

When resistors are connected in parallel, the voltage across each resistor is the same. In this case, the three identical resistors are connected in parallel to a 12-V battery. Therefore, the voltage across any one of the resistors will also be 12 V.

When resistors are connected in parallel, the total resistance is given by the formula 1/Rt = 1/R1 + 1/R2. In this case, 1/Rt = 1/15 + 1/30 = 1/10. Therefore, the total resistance is 10 Ω.

When the combination of resistors is connected in series with the battery and a 20 Ω resistor, the total resistance is the sum of all resistances, which is 10 + 20 = 30 Ω.

Using Ohm's Law, V = IR, we can calculate the current through the circuit. V = 9.0 V and R = 30 Ω, so I = 9.0/30 = 0.30 A.

Since the 15 Ω resistor is in parallel with the combination, it has the same voltage across it as the rest of the circuit, which is 9.0 V. Using Ohm's Law again, we can calculate the current through the 15 Ω resistor. R = 15 Ω and V = 9.0 V, so I = 9.0/15 = 0.20 A.

Therefore, the current through the 15 Ω resistor is 0.20 A.

When two identical storage batteries are connected in series, the positive terminal of one battery is connected to the negative terminal of the other battery. This arrangement creates a closed loop circuit with no voltage difference across the two batteries. As a result, the combination will provide zero volts.

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. This means that as more and more capacitors are added in parallel, the total capacitance of the combination increases. Therefore, the statement "As more and more capacitors are connected in parallel, the equivalent capacitance of the combination increases" is always true.

When a capacitor is discharged through a resistor, the voltage across the capacitor decreases exponentially over time according to the equation V = V0 * e^(-t/RC), where V is the voltage at time t, V0 is the initial voltage, R is the resistance, and C is the capacitance. In this case, the initial voltage is 12 V, the resistance is 4.0 * 10^6 Ω, and the capacitance is 2.0 μF. We need to find the time it takes for the voltage to drop to 3.0 V. Plugging in the values into the equation, we get 3 = 12 * e^(-t/(2.0 * 10^-6 * 4.0 * 10^6)). Solving for t gives us approximately 11 seconds.

When two resistors are connected in series, their resistances add up. In this case, the combination of 2.0 Ω and 4.0 Ω in series will have a total resistance of 6.0 Ω. This combination is then connected in parallel with a 3.0 Ω resistor. When resistors are connected in parallel, the reciprocal of their resistances add up. So, the reciprocal of 6.0 Ω is 1/6.0 Ω, and the reciprocal of 3.0 Ω is 1/3.0 Ω. Adding these reciprocals gives a total of 1/6.0 Ω + 1/3.0 Ω = 1/2.0 Ω. Taking the reciprocal of this total gives an equivalent resistance of 2.0 Ω.