Circuits with Resistors and Inductors - AP Physics C Electricity and Magnetism

A series LR circuit with $V=15\,\text{V}$ , $R=5\,\Omega$ , $L=5\,\text{H}$ closes; Refer to the scenario above. How does the current change over time after the switch is closed?
Answer: $i(t)=\frac{V}{R}\left(1-e^{-tR/L}\right)$ This question tests AP Physics C skills in understanding current growth equations in LR circuits. When a switch closes in an LR circuit, the current grows from zero to a steady-state value of V/R following the equation i(t) = (V/R)(1 - $e^{-tR/L}$ ). In this scenario, with equal values of R = 5 Ω and L = 5 H, the time constant is τ = L/R = 1 s, and the steady-state current is V/R = 15/5 = 3 A. Choice A is correct because it shows the proper form with current approaching V/R and the correct exponential decay term $e^{-tR/L}$ . Choice B is incorrect because it has the exponent as -tL/R instead of -tR/L, which would give τ = R/L instead of the correct τ = L/R. To help students: Check dimensions - the exponent must be dimensionless, so t must be multiplied by something with units of 1/time. Remember that R/L has units of 1/time, while L/R has units of time.
What is the time constant of this circuit after the switch is closed?
Answer: $\frac{L(R_1+R_2)}{R_1R_2}$ 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 $\tau = \frac{L}{R_{eq}} = \frac{L}{\frac{R_1R_2}{R_1+R_2}} = \frac{L(R_1+R_2)}{R_1R_2}$ .
In series $R=8\,\Omega$ , $L=4\,\text{H}$ , switch opens and current drops; Refer to the scenario above. Explain the significance of Lenz's Law in the scenario provided.
Answer: Induced emf opposes current decrease, sustaining current momentarily This question tests AP Physics C skills in understanding Lenz's Law and its application to inductor behavior in circuits. Lenz's Law states that the induced emf in an inductor opposes the change in current that causes it, which is crucial for understanding transient behavior in LR circuits. In this scenario, when the switch opens and current begins to decrease, the inductor generates an emf that opposes this decrease, temporarily sustaining the current flow. Choice A is correct because it accurately describes how the induced emf opposes the current decrease, preventing instantaneous current drop. Choice B is incorrect because it suggests the induced emf aids the decrease, which violates Lenz's Law and would make the current stop instantly. To help students: Remember the mnemonic 'inductors hate change' - they always oppose whatever is happening to the current. Visualize the inductor as trying to maintain the status quo of current flow.
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 $\tau_B$ ?
Answer: $I_A < I_B$ and $\tau_A < \tau_B$ 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, $\tau_A < \tau_B$ .
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.
Answer: Back emf opposes the current drop, creating a voltage spike 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.
At what rate is energy being stored in the magnetic field of the inductor at time $t=0$ ?
Answer: At a rate of zero, because the current is zero. 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 $E/L$ .
What is the approximate current in the circuit at time $t = 0.40$ s?
Answer: 1.26 A 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, $I(0.40 \text{ s}) = (2.0 \text{ A})(1 - e^{-1}) = (2.0 \text{ A})(1 - 0.368) = (2.0 \text{ A})(0.632) \approx 1.26$ A.
How does the potential difference across the resistor, $V_R$ , change with time?
Answer: It decreases exponentially from its initial value of $I_0R$ . 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.
What is the total energy dissipated by the resistor as the current decays from $I_0$ to zero?
Answer: $\frac{1}{2}L I_0^2$ 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.
In a series $R$ – $L$ loop, Kirchhoff gives $V- iR - L\,di/dt=0$ ; induced emf opposes change; Refer to the scenario above. Explain the significance of Lenz's Law in the scenario provided.
Answer: Induced emf opposes the current change This question tests AP Physics C skills in understanding LR circuits and electromagnetic induction principles. Lenz's Law states that the induced EMF in an inductor opposes the change in current that produces it, mathematically expressed as ε = -L(di/dt). In the given Kirchhoff equation V - iR - L(di/dt) = 0, the L(di/dt) term represents this induced EMF opposing current changes. Choice B is correct because Lenz's Law fundamentally means the inductor creates a back-EMF that opposes any attempt to change the current, whether increasing or decreasing. Choice A incorrectly states the EMF reinforces changes, C confuses inductors with capacitors (inductors don't store charge), and D misunderstands that current changes gradually, not instantly. To help students: Remember the negative sign in Faraday's law embodies Lenz's principle of opposition. Visualize the inductor as an 'inertial' element that resists changes in current flow, similar to how mass resists changes in velocity.
If the resistance is doubled to $2R_0$ and the inductance is halved to $L_0/2$ , what is the new time constant $\tau_{new}$ ?
Answer: $\tau_0 / 4$ 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 $\tau_{new} = (L_0/2) / (2R_0) = L_0 / (4R_0) = (1/4)(L_0/R_0) = \tau_0/4$ .
Immediately after the switch is closed (at $t=0^+$ ), what is the potential difference across the inductor?
Answer: $E$ , because the inductor opposes the change in current, creating a back emf equal to the battery's emf. 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 $I=0$ at $t=0^+$ , 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 $E$ .
A series $R=4\,\Omega$ , $L=2\,\text{H}$ is connected to $8\,\text{V}$ at $t=0$ ; Refer to the scenario above. How does the current change over time after the switch is closed?
Answer: $i(t)=2\left(1-e^{-2t}\right)\,\text{A}$ This question tests AP Physics C skills in understanding LR circuits and electromagnetic induction principles. When a switch closes in an RL circuit, the current grows exponentially according to i(t) = (V/R)(1 - $e^{-t/τ}$ ), where τ = L/R is the time constant. In this scenario, V = 8 V, R = 4 Ω, and L = 2 H, giving τ = 2/4 = 0.5 s and steady-state current I₀ = V/R = 8/4 = 2 A. Choice A is correct because i(t) = 2(1 - $e^{-t/0.5}$ ) = 2(1 - $e^{-2t}$ ) A, showing exponential growth toward 2 A. Choice B incorrectly represents decay rather than growth, while C has the wrong exponent, and D has a positive exponent which would cause unbounded growth. To help students: Remember that closing a switch causes current growth (1 - $e^{-t/τ}$ ) while opening causes decay ( $e^{-t/τ}$ ). Always verify that the exponent is negative and equals -t/τ = -tR/L.
What is the physical significance of the time constant, $\tau = L/R$ , for this circuit?
Answer: It is the time required for the current to reach approximately 63% of its final steady-state value. 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 $t=\tau$ .
What is the primary reason for this delay in the current reaching its maximum value?
Answer: The inductor generates a back emf that opposes the increase in current, a property known as inductance. 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.
Which equation correctly represents Kirchhoff's loop rule for this circuit at time $t$ ?
Answer: $E - IR - L \frac{dI}{dt} = 0$ 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 $E - IR - L \frac{dI}{dt} = 0$ .
What is the value of $U_0$ ?
Answer: $\frac{1}{2}L(E/R)^2$ 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 $U_0 = \frac{1}{2}L(E/R)^2$ .
In series $R=7\,\Omega$ , $L=0.7\,\text{H}$ , switch opens causing rapid current drop; Refer to the scenario above. Explain the significance of Lenz's Law in the scenario provided.
Answer: Induced emf opposes change, producing a voltage spike at opening This question tests AP Physics C skills in understanding Lenz's Law and its practical implications in LR circuits. Lenz's Law dictates that an inductor's induced emf opposes changes in current, which can create significant voltage spikes when switches open rapidly. In this scenario, when the switch opens in the R = 7 Ω, L = 0.7 H circuit, the inductor opposes the rapid current decrease by generating a large back-emf. Choice A is correct because it accurately describes how the induced emf opposes the change and can produce dangerous voltage spikes across the opening switch. Choice B is incorrect because the induced emf cannot cancel resistance effects or maintain current indefinitely without an energy source. To help students: Remember that $V_L$ = -L(di/dt), so rapid current changes (large di/dt) create large voltages. This is why spark gaps or diodes are often placed across inductors in practical circuits.
After a very long time ( $t \to \infty$ ), what is the potential difference across the inductor?
Answer: Zero, because the current has reached a constant maximum value, so its rate of change is zero. 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.
Which of the following describes the current $I(t)$ in the resistor for $t > 0$ ?
Answer: The current decays exponentially toward zero with a time constant of $L/R$ . 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 $\tau = L/R$ .

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