AP Physics C Electricity and Magnetism Quiz: Kirchhoffs Loop Rule

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Kirchhoffs Loop RuleQuestion 1 of 20

A circuit contains an ideal battery $\mathcal{E}$ , a resistor $R_1$ , a capacitor $C$ , and an inductor $L$ . The resistor and inductor are in one parallel branch, while the capacitor is in another parallel branch. This parallel combination is in series with the battery. The switch is closed at $t=0$ .

After the switch has been closed for a very long time (in the steady state), which equation correctly describes the circuit based on Kirchhoff's loop rule?

The current in the circuit is zero everywhere.

\mathcal{E} - I R_1 = 0

, where

I

is the current through the inductor branch.

\mathcal{E} - \frac{Q_{max}}{C} = 0

, where

Q_{max}

is the final charge on the capacitor.

\mathcal{E} - I R_1 - \frac{Q_{max}}{C} = 0

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Kirchhoffs Loop Rule

Practice Kirchhoffs Loop Rule 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 Kirchhoffs Loop Rule, 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.

All questions

Question 1

After the switch has been closed for a very long time (in the steady state), which equation correctly describes the circuit based on Kirchhoff's loop rule?

The current in the circuit is zero everywhere.
$\mathcal{E} - I R_1 = 0$ , where $I$ is the current through the inductor branch.
$\mathcal{E} - \frac{Q_{max}}{C} = 0$ , where $Q_{max}$ is the final charge on the capacitor. (correct answer)
$\mathcal{E} - I R_1 - \frac{Q_{max}}{C} = 0$

Explanation: After a very long time, the circuit reaches a steady state. The inductor acts as a short circuit (a wire with zero resistance), so the potential drop across it is zero. The capacitor becomes fully charged and acts as an open circuit, allowing no DC current to flow through its branch. Therefore, the current through the inductor branch becomes zero. The potential difference across the parallel branches must be equal. The potential across the inductor branch is

V_L + V_{R1} = 0 + I R_1 = 0

, so I=0. The loop containing the battery and the capacitor gives

+\mathcal{E} - V_C = 0

, or

\mathcal{E} - Q_{max}/C = 0

Question 2

In a simple series circuit, a $12\,\text{V}$ ideal battery is connected to a $2\,\Omega$ resistor and a $4\,\Omega$ resistor. The negative terminal of the battery is connected to a point defined to have zero potential. A clockwise current flows in the circuit from the positive terminal.

What is the electric potential at the point located between the $2\,\Omega$ and $4\,\Omega$ resistors?

$12\,\text{V}$
$8\,\text{V}$ (correct answer)
$4\,\text{V}$
$2\,\text{V}$

Explanation: First, find the total resistance

R_{tot} = 2\,\Omega + 4\,\Omega = 6\,\Omega

. The current is

I = \mathcal{E}/R_{tot} = 12\,\text{V} / 6\,\Omega = 2\,\text{A}

. Starting from the negative terminal (0 V), the potential at the positive terminal is

+12\,\text{V}

. Traversing through the

2\,\Omega

resistor in the direction of the current, the potential drops by

IR = (2\,\text{A})(2\,\Omega) = 4\,\text{V}

. The potential at the point between the resistors is

12\,\text{V} - 4\,\text{V} = 8\,\text{V}

Question 3

A single-loop circuit contains two ideal batteries and one resistor. Battery 1 has EMF $\mathcal{E}_1 = 12\,\text{V}$ . Battery 2 has EMF $\mathcal{E}_2 = 3\,\text{V}$ . The resistor has resistance $R = 6\,\Omega$ . The batteries are connected in opposition, such that their positive terminals are connected to each other through the resistor.

By applying Kirchhoff's loop rule, what is the magnitude of the current flowing through the resistor?

$1.5\,\text{A}$ (correct answer)
$2.5\,\text{A}$
$2.0\,\text{A}$
$0.5\,\text{A}$

Explanation: Let's assume the current

I

flows in the direction driven by the larger battery,

\mathcal{E}_1

(counter-clockwise). Traversing the loop in this direction, we get a potential gain

+\mathcal{E}_1

, a drop

-IR

, and a drop

- \mathcal{E}_2

(since we traverse it from positive to negative). The loop rule is

+\mathcal{E}_1 - \mathcal{E}_2 - IR = 0

. Solving for

I

I = (\mathcal{E}_1 - \mathcal{E}_2) / R = (12\,\text{V} - 3\,\text{V}) / 6\,\Omega = 9\,\text{V} / 6\,\Omega = 1.5\,\text{A}

Question 4

A student measures the electric potential $V$ at points along a single-loop DC circuit containing an ideal battery and several resistors. They plot $V$ versus position $s$ along the wire, starting from the negative terminal of the battery (where $V=0$ ). The graph shows a vertical jump up at the battery, followed by several segments of linear decrease, returning to zero at the starting point.

Which aspect of this graph is a direct illustration of Kirchhoff's loop rule?

The sum of the magnitudes of the potential drops across the resistors is equal to the battery's EMF. (correct answer)
The slope of the graph is negative in the regions corresponding to resistors.
The potential jump at the battery is positive and instantaneous.
The total current flowing out of the battery is conserved throughout the circuit.

Explanation: Kirchhoff's loop rule states the sum of potential changes around a loop is zero (

\sum \Delta V = 0

). In this circuit, this means

+\mathcal{E} - \sum IR_i = 0

, or

\mathcal{E} = \sum IR_i

. The graph shows the potential gain from the battery (

\mathcal{E}

) is exactly cancelled by the sum of the potential drops across the resistors (

\sum IR_i

) as the potential returns to its starting value. This directly illustrates the rule.

Question 5

A battery with emf $\varepsilon=15\,\text{V}$ and internal resistance $r=0.50\,\Omega$ is connected in series with an external resistor $R=4.5\,\Omega$ . Using the loop rule, what is the potential difference across the battery terminals during operation?

13.5 V (correct answer)
15.0 V
1.50 V
0.135 V

Explanation: This question tests AP Physics C concepts of applying Kirchhoff's Loop Rule in circuit analysis. Kirchhoff's Loop Rule requires that the sum of all voltage changes around a closed loop equals zero. In this circuit, a battery with emf ε=15V and internal resistance r=0.50Ω connects to external resistor R=4.5Ω. Choice A (13.5 V) is correct because the current is I = ε/(R+r) = 15/(4.5+0.5) = 3A, making terminal voltage = ε - Ir = 15 - 3(0.5) = 13.5V. Choice B (15.0 V) is incorrect as it neglects the internal resistance effect. To help students: Reinforce that terminal voltage is always less than emf when current flows. Practice calculating both current and terminal voltage in circuits with internal resistance.

Question 6

A battery with emf $\varepsilon=18\,\text{V}$ and internal resistance $r=2.0\,\Omega$ is connected to an external load resistor $R=7.0\,\Omega$ in series (single loop). Using Kirchhoff's Loop Rule, what is the potential difference across the battery terminals while delivering current to the load?

14.0 V (correct answer)
18.0 V
4.0 V
1.4 V

Explanation: This question tests AP Physics C concepts of applying Kirchhoff's Loop Rule with internal resistance. Kirchhoff's Loop Rule states that the sum of all voltage changes around a closed loop must equal zero. In this circuit, an 18V battery with 2Ω internal resistance connects to a 7Ω load. Applying the loop rule: 18V - I(2Ω) - I(7Ω) = 0, giving I = 18V/9Ω = 2A. The terminal voltage equals the emf minus the internal voltage drop: V_terminal = 18V - (2A)(2Ω) = 14V, making choice A correct. Choice B (18.0 V) is incorrect as it ignores the internal resistance effect. To help students: Emphasize that real batteries have internal resistance affecting terminal voltage. Practice problems involving power delivery and efficiency with internal resistance.

Question 7

An RC circuit consisting of a resistor $R$ and capacitor $C$ is discharging. The loop equation is given by $IR - \frac{Q}{C} = 0$ .

In the context of the discharging process where current $I$ flows away from the positive plate, the current is related to the charge $Q$ by $I = -\frac{dQ}{dt}$ . Which differential equation accurately describes the charge $Q$ on the capacitor?

$-R\frac{dQ}{dt} - \frac{Q}{C} = 0$ (correct answer)
$R\frac{dQ}{dt} - \frac{Q}{C} = 0$
$-R\frac{dQ}{dt} + \frac{Q}{C} = 0$
$R\frac{dQ}{dt} + \frac{Q}{C} = 0$

Explanation: The question presents the loop rule equation as

IR - Q/C = 0

. This implies a traversal direction opposite to the standard one, but it is internally consistent. If we substitute

I = -dQ/dt

into this given equation, we get

(-dQ/dt)R - Q/C = 0

, which is

-R(dQ/dt) - Q/C = 0

. Note that the standard formulation

+Q/C - IR = 0

leads to

+Q/C - (-dQ/dt)R = 0

, or

Q/C + R(dQ/dt) = 0

, which is mathematically equivalent.

Question 8

In the circuit shown, two loops share resistor $R_3$ . The left loop contains a $9\,\text{V}$ battery and $R_1=3\,\Omega$ . The right loop contains a $6\,\text{V}$ battery and $R_2=2\,\Omega$ . The shared resistor is $R_3=4\,\Omega$ . Currents $I_1$ (left loop) and $I_2$ (right loop) are clockwise. Find the current through the branch containing resistor $R_3$ (take downward through $R_3$ as positive), using Kirchhoff's Loop Rule.

Use the diagram for loop identification and sign convention.

0.50 A (correct answer)
-0.50 A
0.05 A
1.25 A

Explanation: This question tests AP Physics C concepts of applying Kirchhoff's Loop Rule in circuit analysis with multiple loops. Kirchhoff's Loop Rule states that the sum of the electric potential differences around any closed loop must be zero, which is crucial for analyzing multi-loop circuits. In this circuit, two loops share resistor R₃, with mesh currents I₁ and I₂ both clockwise. For the left loop: 9V - 3I₁ - 4(I₁-I₂) = 0, giving 9 = 7I₁ - 4I₂. For the right loop: 6V - 2I₂ - 4(I₂-I₁) = 0, giving 6 = -4I₁ + 6I₂. Solving simultaneously: I₁ = 1.5A and I₂ = 1.0A. The current through R₃ (downward positive) is I₁ - I₂ = 0.5A, making choice A correct. Choice B (-0.50 A) would be incorrect as it has the wrong sign. To help students: Practice setting up mesh current equations and maintaining consistent sign conventions. Watch for common errors in handling shared components between loops.

Question 9

A battery with emf $\varepsilon=18\,\text{V}$ and internal resistance $r=2.0\,\Omega$ is connected in series with an external resistor $R=7.0\,\Omega$ to form one closed loop. Using the loop rule, what is the potential difference across the battery terminals while the circuit is operating?

14.0 V (correct answer)
18.0 V
4.00 V
1.40 V

Explanation: This question tests AP Physics C concepts of applying Kirchhoff's Loop Rule in circuit analysis. Kirchhoff's Loop Rule states that the algebraic sum of voltage changes around a closed loop must equal zero. In this circuit, a battery with emf ε=18V and internal resistance r=2.0Ω connects to an external resistor R=7.0Ω. Choice A (14.0 V) is correct because the current is I = ε/(R+r) = 18/(7+2) = 2A, making the terminal voltage = ε - Ir = 18 - 2(2) = 14V. Choice B (18.0 V) is incorrect as it ignores the internal resistance voltage drop. To help students: Stress that terminal voltage equals emf minus the internal voltage drop. Practice problems with varying internal resistance values to build intuition.

Question 10

A circuit consists of an ideal battery with electromotive force $\mathcal{E}$ , a resistor with resistance $R$ , and a capacitor with capacitance $C$ , all connected in series. A clockwise current $I$ is flowing as the capacitor with charge $Q$ is charging.

Which of the following equations correctly represents Kirchhoff's loop rule for this circuit when traversing the loop clockwise, starting from the negative terminal of the battery?

$+\mathcal{E} - IR - \frac{Q}{C} = 0$ (correct answer)
$+\mathcal{E} + IR + \frac{Q}{C} = 0$
$+\mathcal{E} - IR + \frac{Q}{C} = 0$
$-\mathcal{E} - IR - \frac{Q}{C} = 0$

Explanation: Traversing the loop clockwise: the battery provides a potential gain of

+\mathcal{E}

. The resistor causes a potential drop of

-IR

when traversed in the direction of the current. The capacitor, being charged by the clockwise current, has its positive plate encountered first, resulting in a potential drop of

-\frac{Q}{C}

. Summing these potential changes to zero gives

+\mathcal{E} - IR - \frac{Q}{C} = 0

Question 11

A circuit has a top branch with resistor $R_1$ and ideal battery $\mathcal{E}_1$ . It has a middle branch with resistor $R_2$ . It has a bottom branch with resistor $R_3$ and ideal battery $\mathcal{E}_2$ . All three branches are connected in parallel. Current $I_1$ flows to the right through the top branch, and current $I_2$ flows to the right through the middle branch.

Which equation correctly represents Kirchhoff's loop rule for the loop consisting of the top and middle branches, traversed clockwise starting from the leftmost junction?

$+\mathcal{E}_1 - I_1 R_1 - I_2 R_2 = 0$
$+\mathcal{E}_1 - I_1 R_1 + I_2 R_2 = 0$ (correct answer)
$+\mathcal{E}_1 + I_1 R_1 - I_2 R_2 = 0$
$\mathcal{E}_1 - I_1 R_1 = I_2 R_2 - \mathcal{E}_2$

Explanation: Traversing the top branch clockwise (to the right) from the left junction, we go through the battery (

+\mathcal{E}_1

) and the resistor (

-I_1 R_1

). Then, traversing the middle branch clockwise (to the left), we go against the direction of current

I_2

, resulting in a potential gain of

+I_2 R_2

. Summing these gives

+\mathcal{E}_1 - I_1 R_1 + I_2 R_2 = 0

Question 12

A series circuit contains a switch, an ideal battery with EMF $\mathcal{E}$ , a resistor $R$ , and an initially uncharged capacitor $C$ . At time $t=0$ , the switch is closed. Let $Q(t)$ be the charge on the capacitor and $I(t) = dQ/dt$ be the current in the circuit.

Which of the following differential equations is a correct application of Kirchhoff's loop rule to this charging circuit for $t>0$ ?

$\mathcal{E} - R \frac{dQ}{dt} - \frac{Q}{C} = 0$ (correct answer)
$\mathcal{E} - \frac{I}{C} - R\frac{dI}{dt} = 0$
$\mathcal{E} + R \frac{dQ}{dt} + \frac{Q}{C} = 0$
$\mathcal{E} - L\frac{dI}{dt} - \frac{Q}{C} = 0$

Explanation: Applying the loop rule, we sum the potential changes around the circuit. The battery provides a potential gain of

\mathcal{E}

. The resistor causes a potential drop of

IR = R(dQ/dt)

. The capacitor causes a potential drop of

Q/C

. Setting the sum to zero gives

\mathcal{E} - R(dQ/dt) - Q/C = 0

Question 13

A capacitor with capacitance $C$ is initially charged with charge $Q_0$ . At time $t=0$ , it is connected in a simple loop with a resistor of resistance $R$ . Let $Q(t)$ be the charge on the capacitor and $I(t)$ be the current flowing from the positive plate through the resistor.

Which of the following equations correctly represents Kirchhoff's loop rule for this discharging circuit?

$+\frac{Q}{C} - IR = 0$ (correct answer)
$-\frac{Q}{C} - IR = 0$
$+\frac{Q}{C} + IR = 0$
$Q_0/C - I(t)R=0$

Explanation: Starting at the negative plate of the capacitor and traversing through it to the positive plate, there is a potential gain of

+Q/C

. Continuing through the resistor in the direction of the current

I

, there is a potential drop of

-IR

. Summing these potential changes around the loop to zero gives

+Q/C - IR = 0

. Note that for discharging,

I = -dQ/dt

Question 14

A series circuit contains a switch, an ideal battery with EMF $\mathcal{E}$ , a resistor $R$ , and an inductor $L$ . The switch is initially open. At time $t=0$ , the switch is closed. Let $I(t)$ be the current in the circuit.

Which differential equation correctly describes the circuit for $t>0$ based on Kirchhoff's loop rule?

$\mathcal{E} - IR - L\frac{dI}{dt} = 0$ (correct answer)
$\mathcal{E} - IR + L\frac{dI}{dt} = 0$
$\mathcal{E} - I/R - L\frac{dI}{dt} = 0$
$\mathcal{E} + IR + L\frac{dI}{dt} = 0$

Explanation: Applying the loop rule in the direction of the current: the battery provides a potential gain of

\mathcal{E}

. The resistor causes a potential drop of

IR

. As the current is increasing, the inductor generates a back EMF that opposes this change, resulting in a potential drop of

L(dI/dt)

. The sum of potential changes is zero:

\mathcal{E} - IR - L(dI/dt) = 0

Question 15

A non-ideal battery has an EMF $\mathcal{E}$ and an internal resistance $r$ . It is connected to an external resistor of resistance $R$ . The current flowing in the circuit is $I$ .

Which of the following is the correct expression for Kirchhoff's loop rule applied to this circuit, accounting for the internal resistance?

$\mathcal{E} - IR = 0$
$\mathcal{E} - I(R+r) = 0$ (correct answer)
$\mathcal{E} - I(R-r) = 0$
$\mathcal{E} + Ir - IR = 0$

Explanation: The internal resistance

r

acts as a resistor in series with the ideal EMF source. Applying the loop rule, we have a potential gain of

\mathcal{E}

from the ideal source, a potential drop of

-Ir

across the internal resistance, and a potential drop of

-IR

across the external resistor. The equation is

\mathcal{E} - Ir - IR = 0

, which can be rewritten as

\mathcal{E} - I(R+r) = 0

Question 16

A student analyzes a circuit containing an ideal battery $\mathcal{E}$ , a resistor $R$ , and an inductor $L$ in series. They traverse the circuit clockwise, in the direction of an increasing current $I$ , and write the following equation based on Kirchhoff's loop rule: $+\mathcal{E} - IR + L\frac{dI}{dt} = 0$ .

What is the error in the student's equation?

The term for the battery should be negative as it is a source of potential.
The term for the resistor should be positive because it dissipates energy.
The term for the inductor should be negative because its induced EMF opposes the current increase. (correct answer)
The equation should be set equal to the total power, not zero.

Explanation: When traversing a circuit in the direction of an increasing current, the inductor creates a back-EMF that opposes this increase. This back-EMF corresponds to a potential drop. Therefore, the term for the inductor should be

-L(dI/dt)

. The student has the wrong sign for this term.

Question 17

A circuit consists of an ideal battery with EMF $\mathcal{E}$ and a resistor $R_1$ in series with a parallel combination of two resistors, $R_2$ and $R_3$ . Let $I_1$ be the total current from the battery, which splits into $I_2$ and $I_3$ through $R_2$ and $R_3$ , respectively.

Which equation is a valid application of Kirchhoff's loop rule for the loop containing the battery, resistor $R_1$ , and resistor $R_3$ ?

$\mathcal{E} - I_1 R_1 - I_3 R_3 = 0$ (correct answer)
$\mathcal{E} - I_1 R_1 - I_1 R_3 = 0$
$\mathcal{E} - I_2 R_2 - I_3 R_3 = 0$
$\mathcal{E} - I_1(R_1 + R_3) = 0$

Explanation: For the specified loop, we sum the potential changes. Starting from the negative terminal of the battery, we gain

\mathcal{E}

. Then, we traverse

R_1

in the direction of current

I_1

, so the potential drops by

-I_1 R_1

. Finally, we traverse

R_3

in the direction of current

I_3

, so the potential drops by

-I_3 R_3

. Setting the sum to zero gives

\mathcal{E} - I_1 R_1 - I_3 R_3 = 0

Question 18

A circuit has an ideal battery $\mathcal{E}$ in series with a resistor $R_1$ . This combination is connected to a parallel arrangement of a resistor $R_2$ and an inductor $L$ . A switch in the main circuit is closed at $t=0$ . Let $I_1$ be the current through $R_1$ , $I_2$ through $R_2$ , and $I_L$ through $L$ .

Immediately after the switch is closed ( $t=0^+$ ), what is the correct application of Kirchhoff's loop rule for the outer loop containing the battery, $R_1$ , and $R_2$ ?

$\mathcal{E} - I_1(R_1 + R_2) = 0$
$\mathcal{E} - I_1 R_1 = 0$
$\mathcal{E} - L \frac{dI_L}{dt} = 0$
$\mathcal{E} - I_1 R_1 - I_2 R_2 = 0$ (correct answer)

Explanation: Immediately after the switch is closed, the current through the inductor

I_L

must be zero, because current in an inductor cannot change instantaneously. By Kirchhoff's junction rule, the total current

I_1

must therefore pass entirely through resistor

R_2

, so

I_1 = I_2

. The loop rule for the outer loop is

+\mathcal{E} - I_1 R_1 - I_2 R_2 = 0

. This equation is valid at all times, including at

t=0^+

Question 19

For a simple DC circuit containing an ideal battery with EMF $\mathcal{E}$ and a resistor with resistance $R$ , the Kirchhoff loop rule equation is $\mathcal{E} - IR = 0$ . This equation is then multiplied by the current $I$ to yield $\mathcal{E}I - I^2R = 0$ .

What is the primary physical interpretation of the equation $\mathcal{E}I = I^2R$ ?

The net force on the charge carriers is zero, leading to a constant drift velocity.
The total electric potential in the circuit is equal to the total resistance.
The rate at which the battery supplies energy is equal to the rate at which the resistor dissipates energy. (correct answer)
The total energy stored in the battery's electric field equals the energy converted to heat.

Explanation: The term

\mathcal{E}I

represents the power (rate of energy transfer) supplied by the battery. The term

I^2R

represents the power dissipated as heat in the resistor. The equation

\mathcal{E}I = I^2R

is a statement of the conservation of energy applied to the circuit, indicating that the power supplied by the source equals the power consumed by the load.

Question 20

A simple circuit loop contains a resistor of resistance $R$ and an AC generator providing a time-varying EMF given by $\mathcal{E}(t) = \mathcal{E}_0 \cos(\omega t)$ . Let $I(t)$ be the instantaneous current in the circuit.

Which equation correctly applies Kirchhoff's loop rule to this AC circuit?

$\mathcal{E}_0 \cos(\omega t) - I(t)R = 0$ (correct answer)
$\mathcal{E}_0 \cos(\omega t) + I(t)R = 0$
$-\omega\mathcal{E}_0 \sin(\omega t) - \frac{dI}{dt}R = 0$
$\mathcal{E}_0 - I(t)R = 0$

Explanation: Kirchhoff's loop rule applies at any instant in time. At any time

t

, the potential gain from the generator is

\mathcal{E}(t) = \mathcal{E}_0 \cos(\omega t)

. The potential drop across the resistor is

I(t)R

. Summing these potential changes around the loop to zero gives

\mathcal{E}_0 \cos(\omega t) - I(t)R = 0