AP Physics C: Electricity and Magnetism Practice Quiz: Circuits with Resistors and Inductors (LR Circuits)
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 10 questions to check your progress.
Question 1 of 10
All Questions (10)
A) To determine the maximum current the circuit can achieve.
B) To measure the total resistance of the circuit.
C) To describe how quickly the circuit reaches a steady state.
D) To calculate the emf of the battery.
Correct Answer: C
According to the provided content, the time constant τ of a circuit is a measure of how quickly an LR circuit will reach a steady state. It does not directly determine the maximum current, total resistance, or emf.
A) As an open switch, blocking all current.
B) As a resistor with resistance equal to R.
C) As a conducting wire with zero resistance.
D) As a source of emf equal to the battery's.
Correct Answer: C
The provided content states that after a time much greater than the time constant of the circuit, an inductor will behave as a conducting wire with zero resistance. This means the current is steady, and the inductor no longer opposes its change.
A) $\mathcal{E} = IR$
B) $\frac{dI}{dt} = 0$
C) $\mathcal{E} = L\frac{dI}{dt}$
D) $\mathcal{E} = 0$
Correct Answer: C
At the moment the switch is closed (t=0), the current I has not yet had time to build up, so I = 0. Substituting I=0 into Kirchhoff's loop rule equation gives $\mathcal{E}=(0)R+L\frac{dI}{dt}$, which simplifies to $\mathcal{E}=L\frac{dI}{dt}$. This shows that initially, the entire battery emf is across the inductor.
A) A large L and a small R.
B) A large L and a large R.
C) A small L and a small R.
D) A small L and a large R.
Correct Answer: D
A circuit reaches steady state quickly if its time constant, τ, is small. The equation for the time constant is $\tau=\frac{L}{R}$. To minimize τ, one must minimize the numerator (L) and maximize the denominator (R). Therefore, a small L and a large R should be chosen.
A) Zero
B) $\mathcal{E}/R$
C) $\mathcal{E}/L$
D) $\mathcal{E}/(R+L)$
Correct Answer: B
After a long time, the inductor behaves as a conducting wire with zero resistance. The circuit effectively becomes a simple circuit with just the battery and the resistor. According to Kirchhoff's loop rule (or Ohm's Law), the steady-state current is $I = \mathcal{E}/R$. This can also be seen from the equation $\mathcal{E}=IR+L\frac{dI}{dt}$, where in steady state, $\frac{dI}{dt}=0$, leaving $\mathcal{E}=IR$.
A) $\tau_2 = \tau_1$
B) $\tau_2 = 2\tau_1$
C) $\tau_2 = 4\tau_1$
D) $\tau_2 = \frac{1}{4}\tau_1$
Correct Answer: C
The time constant is given by the formula $\tau = \frac{L}{R}$. For Circuit 1, $\tau_1 = \frac{L_1}{R_1}$. For Circuit 2, $\tau_2 = \frac{L_2}{R_2} = \frac{2L_1}{R_1/2} = 4\frac{L_1}{R_1}$. Therefore, $\tau_2 = 4\tau_1$.
A) The potential drop across the resistor.
B) The rate of energy dissipation in the circuit.
C) The potential difference (or induced emf) across the inductor.
D) The total energy stored in the inductor.
Correct Answer: C
In the application of Kirchhoff's loop rule, each term represents a potential difference. $\mathcal{E}$ is the potential gain from the battery, and $IR$ is the potential drop across the resistor. Therefore, the term $L\frac{dI}{dt}$ must represent the potential difference across the remaining component, the inductor.
A) Immediately after the switch is closed (t=0).
B) At a time equal to one time constant, t = τ.
C) A long time after the switch is closed (t → ∞).
D) The potential difference across the resistor is always constant.
Correct Answer: C
The potential difference across the resistor is given by $V_R = IR$. This value is maximum when the current, I, is maximum. The current in an LR circuit builds up over time and reaches its maximum, steady-state value a long time after the switch is closed. At this point, the inductor acts as a wire, and the current is $I_{max} = \mathcal{E}/R$.
A) 50 s
B) 2 s
C) 0.5 s
D) 15 s
Correct Answer: B
The time constant τ is calculated using the formula $\tau = \frac{L}{R}$. Given L = 10 H and R = 5 Ω, the time constant is $\tau = \frac{10 \text{ H}}{5 \text{ } \Omega} = 2 \text{ s}$.
A) A long time after the switch is closed, when the current is constant.
B) Immediately after the switch is closed, when the current is zero.
C) When the current is half its maximum value.
D) The rate of change of current is constant throughout the process.
Correct Answer: B
From the equation $\mathcal{E}=IR+L\frac{dI}{dt}$, we can solve for the rate of change of current: $\frac{dI}{dt} = \frac{\mathcal{E}-IR}{L}$. This value is greatest when the term $IR$ is smallest. The smallest value of $IR$ is zero, which occurs at t=0, immediately after the switch is closed. At this point, $\frac{dI}{dt} = \frac{\mathcal{E}}{L}$, its maximum value.