AP Physics C Electricity and Magnetism Quiz: Compound Direct Current Circuits

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Compound Direct Current CircuitsQuestion 1 of 20

An ideal $12\,\text{V}$ battery is connected to a circuit consisting of a $2\,\Omega$ resistor in series with a parallel combination of a $3\,\Omega$ resistor and a $6\,\Omega$ resistor.

What is the current flowing through the $3\,\Omega$ resistor?

1.0\,\text{A}

1.5\,\text{A}

2.0\,\text{A}

3.0\,\text{A}

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Compound Direct Current Circuits

Practice Compound Direct Current Circuits in AP Physics C Electricity and Magnetism with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Compound Direct Current Circuits, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Electricity and Magnetism.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

An ideal $12\,\text{V}$ battery is connected to a circuit consisting of a $2\,\Omega$ resistor in series with a parallel combination of a $3\,\Omega$ resistor and a $6\,\Omega$ resistor.

What is the current flowing through the $3\,\Omega$ resistor?

$1.0\,\text{A}$
$1.5\,\text{A}$
$2.0\,\text{A}$ (correct answer)
$3.0\,\text{A}$

Explanation: First, find the equivalent resistance of the parallel part:

R_p = (1/3 + 1/6)^{-1} = (3/6)^{-1} = 2\,\Omega

. The total equivalent resistance is

R_{eq} = 2\,\Omega + R_p = 2\,\Omega + 2\,\Omega = 4\,\Omega

. The total current from the battery is

I_{total} = V/R_{eq} = 12\,\text{V} / 4\,\Omega = 3\,\text{A}

. The voltage across the parallel part is

V_p = I_{total} \times R_p = 3\,\text{A} \times 2\,\Omega = 6\,\text{V}

. The current through the

3\,\Omega

resistor is

I_3 = V_p / 3\,\Omega = 6\,\text{V} / 3\,\Omega = 2.0\,\text{A}

Question 2

Four identical resistors, each with resistance $R$ , are available. They are first connected in series to an ideal battery, and the total current is $I_S$ . They are then disconnected and reconnected in parallel to the same battery, and the total current is $I_P$ .

What is the ratio of the parallel current to the series current, $I_P / I_S$ ?

$1/16$
$1/4$
$4$
$16$ (correct answer)

Explanation: When the four resistors are in series, the equivalent resistance is

R_S = R+R+R+R = 4R

. The current is

I_S = \mathcal{E}/(4R)

. When the four resistors are in parallel, the equivalent resistance is

R_P = (1/R + 1/R + 1/R + 1/R)^{-1} = (4/R)^{-1} = R/4

. The current is

I_P = \mathcal{E}/(R/4) = 4\mathcal{E}/R

. The ratio is

I_P / I_S = (4\mathcal{E}/R) / (\mathcal{E}/(4R)) = 16

Question 3

A resistor $R_1 = 4\,\Omega$ is connected in series with a parallel combination of resistors $R_2 = 12\,\Omega$ and $R_3 = 6\,\Omega$ . The circuit is connected to an ideal DC voltage source.

Which of the following correctly compares the power $P_1$ , $P_2$ , and $P_3$ dissipated by the respective resistors?

$P_1 > P_2 > P_3$
$P_1 > P_3 > P_2$ (correct answer)
$P_3 > P_2 > P_1$
$P_2 > P_3 > P_1$

Explanation: Let the total current be

I

. Power in

R_1

P_1 = I^2 R_1 = 4I^2

. The equivalent resistance of the parallel part is

R_p = (1/12 + 1/6)^{-1} = 4\,\Omega

. The voltage across the parallel part is

V_p = I R_p = 4I

. The power in

R_2

P_2 = V_p^2 / R_2 = (4I)^2 / 12 = 16I^2/12 = (4/3)I^2

. The power in

R_3

P_3 = V_p^2 / R_3 = (4I)^2 / 6 = 16I^2/6 = (8/3)I^2

. Comparing the powers:

P_1 = 4I^2

P_2 \approx 1.33I^2

P_3 \approx 2.67I^2

. Therefore, the correct ranking is

P_1 > P_3 > P_2

Question 4

A network of resistors with a total equivalent resistance $R_{ext}$ is connected to a battery with EMF $E$ and internal resistance $r$ . The current flowing is $I$ . If the same resistor network were connected to an ideal battery with the same EMF $E$ , the current would be $I_{ideal}$ . Which expression correctly relates $I_{ideal}$ to $I$ ?

$I_{ideal} = I(1 + r/R_{ext})$ (correct answer)
$I_{ideal} = I(1 - r/R_{ext})$
$I_{ideal} = I(R_{ext}/r)$
$I_{ideal} = I(r/R_{ext})$

Explanation: For the nonideal battery,

I = \mathcal{E} / (R_{ext} + r)

. For the ideal battery,

I_{ideal} = \mathcal{E} / R_{ext}

. From the first equation, we can write

E = I(R_{ext} + r)

. Substituting this into the second equation gives

I_{ideal} = I(R_{ext} + r) / R_{ext}

. Distributing the denominator gives

I_{ideal} = I(R_{ext}/R_{ext} + r/R_{ext}) = I(1 + r/R_{ext})

Question 5

An ideal battery is connected to a circuit where resistor $R_1$ is in series with a parallel combination of resistors $R_2$ and $R_3$ . It is known that $R_2 < R_3$ . The currents through the resistors are $I_1$ , $I_2$ , and $I_3$ , respectively.

Which of the following expressions correctly ranks the magnitudes of the currents?

$I_1 > I_2 > I_3$ (correct answer)
$I_1 > I_3 > I_2$
$I_3 > I_2 > I_1$
$I_2 > I_3 > I_1$

Explanation: The current

I_1

flows through

R_1

and is the total current for the circuit. This current then splits to flow through the parallel branches containing

R_2

and

R_3

. By Kirchhoff's junction rule,

I_1 = I_2 + I_3

, so

I_1

is the largest current. In a parallel combination, the potential difference across each branch is the same. Since

I = V/R

, the branch with the lower resistance will have the higher current. Because

R_2 < R_3

, it follows that

I_2 > I_3

. Therefore, the correct ranking is

I_1 > I_2 > I_3

Question 6

An ideal battery is connected to a $5\,\Omega$ resistor in series with a parallel combination of a $10\,\Omega$ resistor and a $30\,\Omega$ resistor.

An ideal voltmeter is connected in parallel with the $5\,\Omega$ resistor. If the voltmeter reads $15\,\text{V}$ , what is the EMF of the battery?

$15.0\,\text{V}$
$22.5\,\text{V}$
$37.5\,\text{V}$ (correct answer)
$52.5\,\text{V}$

Explanation: The voltmeter reading gives the potential difference across the

5\,\Omega

resistor. The current through this resistor (which is the total current) is

I = V/R = 15\,\text{V} / 5\,\Omega = 3\,\text{A}

. Next, find the equivalent resistance of the parallel branch:

R_p = (1/10 + 1/30)^{-1} = (3/30 + 1/30)^{-1} = (4/30)^{-1} = 7.5\,\Omega

. The potential difference across the parallel branch is

V_p = I R_p = 3\,\text{A} \times 7.5\,\Omega = 22.5\,\text{V}

. The EMF of the battery is the sum of the potential differences across the series components:

EMF = V_5 + V_p = 15\,\text{V} + 22.5\,\text{V} = 37.5\,\text{V}

Question 7

A simple circuit consists of a battery and a resistor $R$ . An ammeter is placed in series with the resistor. Initially, an ideal ammeter is used and measures a current $I_0$ .

If the ideal ammeter is replaced by a real ammeter with a nonzero internal resistance, how will the new current measurement compare to $I_0$ ?

It will be greater than $I_0$ because the real ammeter draws additional power.
It will be less than $I_0$ because the real ammeter increases the total resistance of the circuit. (correct answer)
It will be equal to $I_0$ because ammeters are designed to measure current without altering it.
It could be greater or less than $I_0$ , depending on the battery's internal resistance.

Explanation: A real ammeter has a small but nonzero internal resistance. When placed in series in the circuit, this resistance adds to the existing resistance

R

. This increases the total equivalent resistance of the circuit. According to Ohm's law,

I = V/R_{total}

, an increase in the total resistance will cause a decrease in the total current. Therefore, the measured current will be less than the current measured by the ideal ammeter.

Question 8

A circuit contains an ideal battery connected in series to two resistors, $R_1$ and $R_2$ . An ideal voltmeter connected across $R_1$ measures a potential difference $V_{ideal}$ .

If the ideal voltmeter is replaced by a real voltmeter with a large but finite resistance, how will the measured potential difference $V_{real}$ compare to $V_{ideal}$ ?

$V_{real} < V_{ideal}$ because the voltmeter reduces the equivalent resistance of the part of the circuit it is connected to. (correct answer)
$V_{real} > V_{ideal}$ because the voltmeter increases the total current drawn from the battery.
$V_{real} = V_{ideal}$ because a real voltmeter's resistance is high enough to be considered infinite.
$V_{real} < V_{ideal}$ because the voltmeter adds its resistance in series with $R_1$ , drawing less current.

Explanation: A real voltmeter is connected in parallel with resistor

R_1

. This creates a parallel combination with an equivalent resistance

R_{p1}

that is less than

R_1

. This decrease in resistance for one part of the series circuit leads to a decrease in the total equivalent resistance of the entire circuit. According to the voltage divider principle, the fraction of the total voltage that drops across the

R_1

-voltmeter combination will be smaller than the fraction that dropped across

R_1

alone. Thus,

V_{real} < V_{ideal}

Question 9

A resistor with resistance $R_1$ is connected in series with a parallel combination of two other resistors with resistances $R_2$ and $R_3$ . Which of the following expressions represents the equivalent resistance of this entire network?

$R_1 + R_2 + R_3$
$R_1 + \frac{R_2 R_3}{R_2 + R_3}$ (correct answer)
$(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3})^{-1}$
$\frac{(R_1 + R_2) R_3}{R_1 + R_2 + R_3}$

Explanation: The equivalent resistance of the parallel combination of

R_2

and

R_3

R_p = (\frac{1}{R_2} + \frac{1}{R_3})^{-1} = \frac{R_2 R_3}{R_2 + R_3}

. This combination is in series with

R_1

, so the total equivalent resistance is found by adding their resistances:

R_{eq} = R_1 + R_p = R_1 + \frac{R_2 R_3}{R_2 + R_3}

Question 10

A $12\,\text{V}$ battery is connected to a $2\,\Omega$ resistor which is in series with a parallel combination of two $4\,\Omega$ resistors.

What is the potential difference across the parallel combination of the two $4\,\Omega$ resistors?

$6\,\text{V}$ (correct answer)
$4\,\text{V}$
$8\,\text{V}$
$12\,\text{V}$

Explanation: The equivalent resistance of two

4\,\Omega

resistors in parallel is

R_p = (1/4 + 1/4)^{-1} = 2\,\Omega

. The total circuit resistance is the sum of the series resistor and this parallel equivalent:

R_{eq} = 2\,\Omega + 2\,\Omega = 4\,\Omega

. The total current is

I = V/R_{eq} = 12\,\text{V} / 4\,\Omega = 3\,\text{A}

. The potential difference across the parallel combination is

V_p = I \times R_p = 3\,\text{A} \times 2\,\Omega = 6\,\text{V}

Question 11

A battery with an electromotive force (EMF) of $9.0\,\text{V}$ and an internal resistance of $0.5\,\Omega$ is connected to a network of resistors. The network consists of a $2.0\,\Omega$ resistor in series with a parallel combination of a $3.0\,\Omega$ resistor and a $6.0\,\Omega$ resistor.

What is the terminal voltage of the battery when connected to this circuit?

$7.5\,\text{V}$
$8.0\,\text{V}$ (correct answer)
$9.0\,\text{V}$
$10.5\,\text{V}$

Explanation: First, calculate the equivalent resistance of the external circuit. The parallel part is

R_p = (1/3 + 1/6)^{-1} = 2.0\,\Omega

. The total external resistance is

R_{ext} = 2.0\,\Omega + R_p = 2.0\,\Omega + 2.0\,\Omega = 4.0\,\Omega

. The total resistance of the entire circuit is

R_{total} = R_{ext} + r = 4.0\,\Omega + 0.5\,\Omega = 4.5\,\Omega

. The current is

I = \mathcal{E} / R_{total} = 9.0\,\text{V} / 4.5\,\Omega = 2.0\,\text{A}

. The terminal voltage is

V_T = \mathcal{E} - Ir = 9.0\,\text{V} - (2.0\,\text{A})(0.5\,\Omega) = 8.0\,\text{V}

Question 12

A circuit contains an ideal battery connected to a resistor $R_1$ in series with another resistor $R_2$ . A third resistor, $R_3$ , is then connected in parallel with $R_2$ .

How does the addition of $R_3$ affect the potential difference across $R_1$ ?

It increases, because the total current from the battery increases. (correct answer)
It decreases, because the total resistance of the circuit decreases.
It remains the same, because $R_1$ and the battery EMF are unchanged.
It increases, because the voltage across the parallel branch must decrease.

Explanation: Adding

R_3

in parallel with

R_2

decreases the equivalent resistance of that branch. This, in turn, decreases the total equivalent resistance of the entire circuit. According to Ohm's law for the whole circuit, a lower total resistance results in a higher total current drawn from the battery. Since the potential difference across

R_1

is given by

V_1 = I_{total}R_1

, an increase in the total current leads to an increase in the potential difference across

R_1

Question 13

A circuit initially consists of two resistors, $R_1$ and $R_2$ , connected in parallel to an ideal battery. A third resistor, $R_3$ , is then added in series with the battery and the parallel combination.

What is the effect of adding resistor $R_3$ on the total power dissipated by the circuit?

The total power increases because there are more resistors in the circuit.
The total power decreases because the total equivalent resistance of the circuit increases. (correct answer)
The total power remains the same because the battery's voltage is constant.
The effect on total power depends on whether $R_3$ is larger or smaller than the parallel combination.

Explanation: Adding a resistor

R_3

in series increases the total equivalent resistance of the circuit. The total power dissipated by the circuit is given by

P = V^2 / R_{eq}

, where

V

is the constant voltage of the battery. Since

R_{eq}

increases, the total power dissipated by the circuit must decrease.

Question 14

A $20\,\text{V}$ ideal battery is connected to a $4\,\Omega$ resistor in series with a parallel combination of a $12\,\Omega$ resistor and a $6\,\Omega$ resistor. An ideal ammeter is connected in series with the $6\,\Omega$ resistor.

What is the approximate reading on the ideal ammeter?

$1.0\,\text{A}$
$1.7\,\text{A}$ (correct answer)
$2.5\,\text{A}$
$3.3\,\text{A}$

Explanation: First, calculate the equivalent resistance of the parallel branch:

R_p = (1/12 + 1/6)^{-1} = (1/12 + 2/12)^{-1} = (3/12)^{-1} = 4\,\Omega

. The total circuit resistance is

R_{eq} = 4\,\Omega + R_p = 4\,\Omega + 4\,\Omega = 8\,\Omega

. The total current is

I_{total} = V/R_{eq} = 20\,\text{V}/8\,\Omega = 2.5\,\text{A}

. The voltage across the parallel branch is

V_p = I_{total}R_p = 2.5\,\text{A} \times 4\,\Omega = 10\,\text{V}

. The current through the

6\,\Omega

resistor is

I_6 = V_p/6\,\Omega = 10\,\text{V}/6\,\Omega \approx 1.67\,\text{A}

Question 15

An ideal $24\,\text{V}$ source supplies a total current of $4.0\,\text{A}$ to a circuit. The circuit consists of a $3.0\,\Omega$ resistor in series with a parallel arrangement of a $6.0\,\Omega$ resistor and an unknown resistor $R_x$ .

What is the resistance of $R_x$ ?

$3.0\,\Omega$
$6.0\,\Omega$ (correct answer)
$9.0\,\Omega$
$12.0\,\Omega$

Explanation: The total equivalent resistance of the circuit is

R_{eq} = V/I = 24\,\text{V} / 4.0\,\text{A} = 6.0\,\Omega

. This total resistance is the sum of the series resistor and the equivalent resistance of the parallel part,

R_p

. So,

R_{eq} = 3.0\,\Omega + R_p

, which gives

R_p = 6.0\,\Omega - 3.0\,\Omega = 3.0\,\Omega

. For the parallel combination,

1/R_p = 1/6.0 + 1/R_x

. Substituting

R_p = 3.0\,\Omega

, we get

1/3.0 = 1/6.0 + 1/R_x

. This gives

1/R_x = 1/3.0 - 1/6.0 = 2/6.0 - 1/6.0 = 1/6.0

. Therefore,

R_x = 6.0\,\Omega

Question 16

A circuit consists of an ideal battery and three identical resistors. Two of the resistors are connected in parallel, and this combination is connected in series with the third resistor.

If a wire of negligible resistance is connected across the two terminals of the series resistor, effectively shorting it, what happens to the total power delivered by the battery?

It decreases, because one resistor is removed from dissipating power.
It remains the same, because the battery's voltage does not change.
It increases, because the total equivalent resistance of the circuit decreases. (correct answer)
It becomes zero, because the current is diverted through the shorting wire.

Explanation: Let each resistor have resistance

R

. The initial equivalent resistance is

R_{eq,1} = R + (1/R + 1/R)^{-1} = R + R/2 = 1.5R

. When the series resistor is shorted, it is bypassed, and the new equivalent resistance is just that of the parallel pair,

R_{eq,2} = R/2

. Since the total equivalent resistance decreases, the total current from the battery (

I=V/R_{eq}

) increases. The total power delivered by the battery is

P = V^2/R_{eq}

. Since

R_{eq}

decreases, the total power

P

increases.

Question 17

A circuit consists of a battery connected to a resistor $R_1$ in series with a parallel combination of resistors $R_2$ and $R_3$ .

If resistor $R_2$ is removed from the circuit, creating an open in its branch, what is the effect on the current passing through resistor $R_1$ ?

It increases, because the parallel resistance is removed.
It decreases, because the total equivalent resistance of the circuit increases. (correct answer)
It remains the same, because $R_1$ is in series with the battery.
It becomes zero, because the circuit is now open.

Explanation: Initially, the total resistance includes the parallel combination of

R_2

and

R_3

. The equivalent resistance of this parallel part is always less than either individual resistance. When

R_2

is removed, the parallel combination is replaced by just

R_3

. Since

R_3

is greater than the equivalent resistance of the original parallel pair, the total equivalent resistance of the circuit increases. According to Ohm's Law (

I=V/R_{eq}

), an increase in total resistance leads to a decrease in the total current. Since

R_1

is in the main branch, the current through it is the total current, which decreases.

Question 18

Five resistors are connected as follows: a $3\,\Omega$ resistor and a $7\,\Omega$ resistor are connected in series to form Branch A. A $4\,\Omega$ resistor and a $6\,\Omega$ resistor are connected in series to form Branch B. Branch A and Branch B are connected in parallel. This combination is then connected in series with a $5\,\Omega$ resistor.

What is the equivalent resistance of this entire network of resistors?

$5.0\,\Omega$
$10.0\,\Omega$ (correct answer)
$15.0\,\Omega$
$25.0\,\Omega$

Explanation: First, find the resistance of each series branch. Branch A:

R_A = 3\,\Omega + 7\,\Omega = 10\,\Omega

. Branch B:

R_B = 4\,\Omega + 6\,\Omega = 10\,\Omega

. Next, find the equivalent resistance of these two branches in parallel:

R_{AB} = (1/10 + 1/10)^{-1} = (2/10)^{-1} = 5\,\Omega

. Finally, this parallel combination is in series with the

5\,\Omega

resistor, so the total equivalent resistance is

R_{eq} = R_{AB} + 5\,\Omega = 5\,\Omega + 5\,\Omega = 10.0\,\Omega

Question 19

A nonideal battery with EMF $E$ and internal resistance $r$ is connected to a variable external resistor $R$ . The power delivered to the external resistor is maximized when which of the following conditions is met?

The external resistance is equal to the internal resistance, $R=r$ . (correct answer)
The external resistance is as small as possible, $R \to 0$ .
The external resistance is as large as possible, $R \to \infty$ .
The external resistance is twice the internal resistance, $R=2r$ .

Explanation: The current in the circuit is

I = \mathcal{E} / (R+r)

. The power delivered to the external resistor is

P = I^2 R = (\frac{\mathcal{E}}{R+r})^2 R

. To find the maximum power, we take the derivative of

P

with respect to

R

and set it to zero. The calculation shows that

dP/dR = 0

when

R=r

. This is the condition for maximum power transfer.

Question 20

A circuit consists of an ideal battery connected to a resistor $R_1$ in series with a parallel combination of two other resistors, $R_2$ and $R_3$ .

If the resistance of $R_2$ is decreased, what is the effect on the potential difference across $R_3$ ?

It decreases, because the total resistance of the circuit decreases. (correct answer)
It increases, because the total current from the battery increases.
It decreases, because a smaller fraction of the total current will flow through $R_3$ .
It remains the same, because $R_3$ and the battery are unchanged.

Explanation: Decreasing

R_2

decreases the equivalent resistance of the parallel combination (

R_p

). This, in turn, decreases the total equivalent resistance of the circuit (

R_{eq} = R_1 + R_p

). A lower total resistance causes the total current from the battery (

I_{total}

) to increase. The potential difference across

R_1

V_1 = I_{total}R_1

, so

V_1

increases. The potential difference across the parallel part (and thus across

R_3

) is given by Kirchhoff's loop rule:

V_p = V_3 = \mathcal{E} - V_1

. Since

V_1

increases,

V_3

must decrease.