AP Physics C Electricity and Magnetism Quiz: Circuits With Resistors And Inductors

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Circuits With Resistors And InductorsQuestion 1 of 20

A series circuit consists of an ideal battery with emf $E$ , a resistor of resistance $R$ , an inductor of inductance $L$ , and an open switch. At time $t=0$ , the switch is closed.

What is the physical significance of the time constant, $\tau = L/R$ , for this circuit?

It is the time required for the current to reach its maximum steady-state value.

It is the time at which the potential difference across the inductor is equal to the potential difference across the resistor.

It is the time required for the current to reach approximately 63% of its final steady-state value.

It is the time at which the energy stored in the inductor is half of its maximum value.

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Circuits With Resistors And Inductors

Practice Circuits With Resistors And Inductors 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 Circuits With Resistors And Inductors, 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

A series circuit consists of an ideal battery with emf $E$ , a resistor of resistance $R$ , an inductor of inductance $L$ , and an open switch. At time $t=0$ , the switch is closed.

What is the physical significance of the time constant, $\tau = L/R$ , for this circuit?

It is the time required for the current to reach its maximum steady-state value.
It is the time at which the potential difference across the inductor is equal to the potential difference across the resistor.
It is the time required for the current to reach approximately 63% of its final steady-state value. (correct answer)
It is the time at which the energy stored in the inductor is half of its maximum value.

Explanation: The time constant

\tau = L/R

in a charging LR circuit is defined as the time it takes for the current to rise to

1 - 1/e

of its final value, which is approximately 63%. The current approaches its maximum value asymptotically and never truly reaches it in finite time. The other conditions occur at different times, not necessarily at

t=\tau

Question 2

A student observes that when a switch is closed in a circuit containing a large inductor and a resistor, the current does not instantaneously jump to its final steady-state value.

What is the primary reason for this delay in the current reaching its maximum value?

The resistor dissipates energy, which slows down the flow of charge carriers through the circuit.
The inductor generates a back emf that opposes the increase in current, a property known as inductance. (correct answer)
The internal resistance of the battery limits the initial rate at which charge can be supplied to the circuit.
The capacitance of the connecting wires stores charge, which must be built up before current can flow steadily.

Explanation: The defining property of an inductor is its inductance, which is its tendency to oppose a change in current. When the switch is closed, the current tries to increase from zero. This change in current creates a changing magnetic flux in the inductor, which in turn induces a back emf that opposes the current's increase. This opposition is what causes the gradual, exponential rise in current rather than an instantaneous one.

Question 3

An LR circuit has been connected to a battery for a long time, and a steady current $I_0$ flows through it. The circuit consists of an inductor $L$ and a resistor $R$ . At $t=0$ , the battery is removed and the inductor and resistor are connected directly together to form a new closed loop.

Which of the following describes the current $I(t)$ in the resistor for $t > 0$ ?

The current remains constant at $I_0$ because the inductor resists any change.
The current immediately drops to zero as the energy source is removed.
The current decays exponentially toward zero with a time constant of $L/R$ . (correct answer)
The current decays linearly to zero over a time interval of $L/R$ .

Explanation: When the battery is removed, the inductor's stored magnetic energy becomes the source for the circuit. The inductor drives a current that initially has the value

I_0

. This energy is dissipated in the resistor, causing the current to decrease. The decay follows an exponential function

I(t) = I_0 e^{-t/\tau}

, where the time constant is

\tau = L/R

Question 4

An inductor with inductance $L$ and a resistor with resistance $R$ are connected in series to a battery with emf $E$ through a switch. The switch is closed at $t=0$ .

At what rate is energy being stored in the magnetic field of the inductor at time $t=0$ ?

At a rate of zero, because the current is zero. (correct answer)
At a rate of $E^2/R$ , which is the maximum power delivered by the battery.
At a rate of $E^2/L$ , derived from the initial rate of current change.
At an infinite rate, because the change in energy is instantaneous at the start.

Explanation: The energy stored in an inductor is

U_L = \frac{1}{2}LI^2

. The rate at which energy is stored is the power,

P_L = \frac{dU_L}{dt} = \frac{d}{dt}(\frac{1}{2}LI^2) = LI \frac{dI}{dt}

. At

t=0

, the current

I

is zero. Therefore, the rate of energy storage

P_L

is also zero, even though

dI/dt

is at its maximum value of

E/L

Question 5

A series circuit consists of an ideal battery with emf $E$ , a resistor of resistance $R$ , an inductor of inductance $L$ , and a switch. The switch is closed at $t=0$ and the circuit is allowed to reach a steady state.

After a very long time ( $t \to \infty$ ), what is the potential difference across the inductor?

Zero, because the current has reached a constant maximum value, so its rate of change is zero. (correct answer)
$E/2$ , because the inductor's impedance matches the resistor's resistance at steady state.
$E$ , because the inductor stores the full energy provided by the battery's emf.
It depends on the value of $L/R$ , which determines the final state of the inductor.

Explanation: After a long time, the circuit reaches a steady state where the current is constant at its maximum value,

I = E/R

. The potential difference across the inductor is given by

V_L = L \frac{dI}{dt}

. Since the current is constant,

\frac{dI}{dt} = 0

, and therefore the potential difference across the inductor is zero. The inductor behaves like a simple connecting wire in a DC circuit at steady state.

Question 6

An inductor with inductance $L$ and a resistor with resistance $R$ are connected in series with a battery. The circuit reaches a steady-state current $I_0$ . The battery is then removed, and the inductor and resistor are connected in a loop to allow the current to decay.

What is the total energy dissipated by the resistor as the current decays from $I_0$ to zero?

$\frac{1}{2}L I_0^2$ (correct answer)
$\frac{1}{2}R I_0^2$
$L R I_0$
Zero, because the energy returns to the inductor.

Explanation: Initially, at steady state, the inductor stores an amount of energy equal to

U_L = \frac{1}{2}L I_0^2

. When the battery is removed, this stored energy is the only energy in the circuit. As the current decays, this energy is entirely dissipated as heat in the resistor. By conservation of energy, the total energy dissipated by the resistor must equal the initial energy stored in the inductor.

Question 7

A circuit contains a battery, a switch, an inductor $L$ , and two resistors $R_1$ and $R_2$ connected in parallel with each other. This parallel combination is in series with the inductor, switch, and battery.

What is the time constant of this circuit after the switch is closed?

$\frac{L}{R_1 + R_2}$
$L \left(\frac{1}{R_1} + \frac{1}{R_2}\right)$
$\frac{L(R_1+R_2)}{R_1R_2}$ (correct answer)
$\frac{L R_1 R_2}{R_1+R_2}$

Explanation: The time constant of an LR circuit is given by

\tau = L/R_{eq}

, where

R_{eq}

is the equivalent resistance seen by the inductor. In this circuit, the two resistors

R_1

and

R_2

are in parallel. Their equivalent resistance is

R_{eq} = \left(\frac{1}{R_1} + \frac{1}{R_2}\right)^{-1} = \frac{R_1R_2}{R_1+R_2}

. Therefore, the time constant is

\tau = \frac{L}{R_{eq}} = \frac{L}{\frac{R_1R_2}{R_1+R_2}} = \frac{L(R_1+R_2)}{R_1R_2}

Question 8

An LR series circuit with a battery and switch is being analyzed. The inductance of the inductor is $L_0$ and the resistance of the resistor is $R_0$ . The time constant is $\tau_0$ .

If the resistance is doubled to $2R_0$ and the inductance is halved to $L_0/2$ , what is the new time constant $\tau_{new}$ ?

$\tau_0 / 4$ (correct answer)
$\tau_0 / 2$
$\tau_0$
$4\tau_0$

Explanation: The time constant of an LR circuit is given by the formula

\tau = L/R

. The initial time constant is

\tau_0 = L_0/R_0

. The new time constant is

\tau_{new} = (L_0/2) / (2R_0) = L_0 / (4R_0) = (1/4)(L_0/R_0) = \tau_0/4

Question 9

A battery of emf $E$ is connected to a switch, an inductor $L$ , and two resistors, $R_1$ and $R_2$ . The inductor $L$ is in series with resistor $R_1$ . This combination is in parallel with resistor $R_2$ . The switch is in series with the battery, controlling the entire circuit.

When determining the time constant for the decay of current through the inductor after the battery has been disconnected for a long time (by shorting the terminals where the battery was), what is the effective resistance used in the calculation of $\tau$ ?

$R_1$ , because it is in the same branch as the inductor.
$R_2$ , because it provides the alternate path for the current.
$R_1 + R_2$ , because the current from the inductor flows through both resistors. (correct answer)
$\frac{R_1 R_2}{R_1 + R_2}$ , because the resistors are in parallel with respect to the inductor during decay.

Explanation: When the battery is removed and its terminals are shorted, the inductor and resistors form a single loop for the current to decay. The current from the inductor must flow through both

R_1

and

R_2

in series to complete the loop. Therefore, the equivalent resistance that governs the decay is the series combination,

R_{eq} = R_1 + R_2

. The time constant would be

\tau = L/(R_1 + R_2)

Question 10

A DC motor with winding $L=40\,\text{mH}$ and $R=2\,\Omega$ is switched off; current was $3\,\text{A}$ ; Refer to the scenario above. Explain the significance of Lenz's Law in the scenario provided.

Back emf opposes the current drop, creating a voltage spike (correct answer)
Back emf aids the current drop, eliminating any voltage spike
Inductor behaves like a capacitor, storing charge to stop arcing
Kirchhoff's loop rule fails, so voltage can't be predicted

Explanation: This question tests AP Physics C skills in understanding LR circuits and electromagnetic induction principles. When a DC motor (essentially an RL circuit) is switched off, the current cannot instantly drop to zero due to the inductor's stored magnetic energy. The inductor generates a back-EMF ε = -L(di/dt) that opposes the rapid current decrease, potentially creating a large voltage spike. Choice A is correct because Lenz's Law dictates that the induced EMF opposes the current drop, maintaining current flow momentarily and creating a voltage spike that can cause arcing at switch contacts. Choice B incorrectly states the EMF aids the drop, C confuses inductors with capacitors, and D wrongly claims Kirchhoff's laws fail. To help students: Calculate the initial spike voltage as V = L(di/dt) ≈ L(I₀/Δt) which can be hundreds of volts for rapid switching. This is why snubber circuits or flyback diodes are used to protect switches from inductive kickback.

Question 11

An $LR$ circuit with $R=8\,\Omega$ and $L=0.40\,\text{H}$ is energized; Refer to the scenario above. What is the time constant of the circuit described?

$\tau=0.050\,\text{s}$ (correct answer)
$\tau=3.2\,\text{s}$
$\tau=20\,\text{s}$
$\tau=0.80\,\text{s}$

Explanation: This question tests AP Physics C skills in understanding LR circuits and electromagnetic induction principles. LR circuits are characterized by a time constant τ = L/R, which determines how quickly current changes in response to voltage changes. In this scenario, the circuit has L = 0.40 H and R = 8 Ω, allowing direct calculation of the time constant. Choice A is correct because τ = L/R = 0.40 H / 8 Ω = 0.05 s = 0.050 s, showing proper application of the time constant formula. Choice C is incorrect as it appears to result from inverting the formula (using R/L instead of L/R), giving 8/0.40 = 20 s. To help students: Always double-check the time constant formula τ = L/R (not R/L) and verify units - henries divided by ohms gives seconds. Practice quick mental calculations by remembering that smaller L or larger R means faster response (smaller τ).

Question 12

Circuit A consists of an inductor $L$ and a resistor $R$ . Circuit B consists of an inductor $2L$ and a resistor $R/2$ . Both are connected to identical batteries at $t=0$ . How does the final steady-state current $I_A$ in circuit A compare to $I_B$ in circuit B, and how does the time constant $\tau_A$ compare to $\tau_B$ ?

$I_A > I_B$ and $\tau_A < \tau_B$
$I_A < I_B$ and $\tau_A < \tau_B$ (correct answer)
$I_A > I_B$ and $\tau_A > \tau_B$
$I_A < I_B$ and $\tau_A > \tau_B$

Explanation: The final steady-state current is determined by

I = E/R

. For circuit A,

I_A = E/R

. For circuit B,

I_B = E/(R/2) = 2E/R

. Thus,

I_A < I_B

. The time constant is

\tau = L/R

. For circuit A,

\tau_A = L/R

. For circuit B,

\tau_B = (2L)/(R/2) = 4L/R

. Thus,

\tau_A < \tau_B

Question 13

A series circuit consists of an ideal battery with emf $E$ , a resistor of resistance $R$ , an inductor of inductance $L$ , and an open switch. At time $t=0$ , the switch is closed.

Immediately after the switch is closed (at $t=0^+$ ), what is the potential difference across the inductor?

Zero, because the current in the circuit is zero.
$E/2$ , because the potential is equally divided between the resistor and inductor.
$E$ , because the inductor opposes the change in current, creating a back emf equal to the battery's emf. (correct answer)
Infinite, because the rate of change of current is momentarily infinite.

Explanation: Immediately after the switch is closed, the current is zero, but it is changing. The inductor opposes this change by inducing a back emf. According to Kirchhoff's loop rule,

E - V_R - V_L = 0

. Since the current

I=0

t=0^+

, the voltage across the resistor

V_R = IR = 0

. Therefore, the voltage across the inductor

V_L

must be equal to the battery's emf

E

Question 14

A series circuit contains an ideal battery of emf $E$ , a resistor $R$ , an inductor $L$ , and a switch. The switch is closed at $t=0$ . At some time $t > 0$ , the current in the circuit is $I$ and the rate of change of current is $dI/dt$ .

Which equation correctly represents Kirchhoff's loop rule for this circuit at time $t$ ?

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

Explanation: According to Kirchhoff's loop rule, the sum of potential changes around a closed loop is zero. Starting from the negative terminal of the battery and moving clockwise: there is a potential gain of

+E

across the battery, a potential drop of

-IR

across the resistor, and a potential drop of

-L \frac{dI}{dt}

across the inductor (since it opposes the increasing current). Thus, the equation is

E - IR - L \frac{dI}{dt} = 0

Question 15

A series circuit consists of a 12 V battery, a 6.0 H inductor, a 3.0 $\Omega$ resistor, and a switch, all initially open.

At the instant the current in the circuit is 2.0 A, what is the potential difference across the inductor?

0 V
6.0 V (correct answer)
12 V
18 V

Explanation: Using Kirchhoff's loop rule for the circuit:

E - V_R - V_L = 0

. We are given

E = 12

V and

I = 2.0

A. The voltage across the resistor is

V_R = IR = (2.0 \text{ A})(3.0 \text{ } \Omega) = 6.0

V. Therefore, the voltage across the inductor is

V_L = E - V_R = 12 \text{ V} - 6.0 \text{ V} = 6.0

Question 16

A circuit contains an ideal battery with emf $E$ , a resistor with resistance $R$ , and an inductor with inductance $L$ . After the circuit has been connected for a long time, the energy stored in the inductor is $U_0$ .

What is the value of $U_0$ ?

$\frac{1}{2}L(E/R)^2$ (correct answer)
$\frac{1}{2}CE^2$
$E^2/R$
$\frac{1}{2}(L/R)E^2$

Explanation: After a long time, the circuit reaches a steady state. The current is constant and has a value of

I_{max} = E/R

. The energy stored in the inductor is given by the formula

U_L = \frac{1}{2}LI^2

. Substituting the steady-state current, we get

U_0 = \frac{1}{2}L(E/R)^2

Question 17

In a simple LR circuit, a battery, resistor, and inductor are in series. The time constant for the current to build up is $\tau$ .

How long does it take for the potential difference across the resistor to reach approximately 63% of its final value?

It takes a time of $\tau$ . (correct answer)
It takes a time of $2\tau$ .
It takes a time less than $\tau$ .
It happens instantaneously at $t=0$ .

Explanation: The potential difference across the resistor is

V_R = IR

. Since

V_R

is directly proportional to the current

I

, it will reach 63% of its final value at the same time the current reaches 63% of its final value. This occurs at one time constant,

t=\tau

. The final value of

V_R

I_{max}R = (E/R)R = E

Question 18

An inductor $L$ and a resistor $R$ are in a circuit that is discharging. The initial current at $t=0$ is $I_0$ .

How does the potential difference across the resistor, $V_R$ , change with time?

It remains constant at $I_0R$ until all energy is dissipated.
It increases exponentially from zero to a maximum value.
It decreases exponentially from its initial value of $I_0R$ . (correct answer)
It decreases linearly from an initial value of $I_0R$ .

Explanation: During discharge, the current is given by

I(t) = I_0 e^{-t/\tau}

. The potential difference across the resistor is

V_R(t) = I(t)R = (I_0 e^{-t/\tau})R = (I_0R)e^{-t/\tau}

. This shows that the voltage across the resistor starts at an initial value of

I_0R

and decays exponentially to zero.

Question 19

An inductor $L$ with initial current $I_0$ is connected in a simple loop with a resistor $R$ at time $t=0$ .

Which of the following differential equations, derived from Kirchhoff's loop rule, describes the current $I$ in the circuit for $t > 0$ ?

$L \frac{dI}{dt} + IR = I_0R$
$- L \frac{dI}{dt} + IR = 0$
$L \frac{dI}{dt} + IR = 0$ (correct answer)
$L \frac{dI}{dt} - IR = 0$

Explanation: In a discharging LR circuit, there is no battery. The inductor acts as the source of emf. Applying Kirchhoff's loop rule, the sum of potential changes is zero. The potential drop across the resistor is

-IR

. The potential change across the inductor is

-L \frac{dI}{dt}

(where

dI/dt

is negative, making the term positive, representing an emf). So the loop equation is

-IR - L \frac{dI}{dt} = 0

, which is equivalent to

L \frac{dI}{dt} + IR = 0

Question 20

A series circuit consists of a 10 V battery, a 2.0 H inductor, a 5.0 $\Omega$ resistor, and a switch. The switch is closed at $t=0$ .

What is the approximate current in the circuit at time $t = 0.40$ s?

0 A
1.26 A (correct answer)
1.73 A
2.0 A

Explanation: The current in a charging LR circuit is given by

I(t) = \frac{E}{R}(1 - e^{-t/\tau})

. The time constant is

\tau = L/R = 2.0 \text{ H} / 5.0 \text{ } \Omega = 0.40

s. The final current is

I_{max} = E/R = 10 \text{ V} / 5.0 \text{ } \Omega = 2.0

A. At

t=0.40

s, which is one time constant, the exponent is

-t/\tau = -1

. So,

I(0.40 \text{ s}) = (2.0 \text{ A})(1 - e^{-1}) = (2.0 \text{ A})(1 - 0.368) = (2.0 \text{ A})(0.632) \approx 1.26