In the circuit shown, an ideal battery with potential difference $V_0$ is connected to three resistors ($R_1 = R$, $R_2 = 2R$, $R_3 = R$), an ideal inductor $L$, and a switch $S$. The switch $S$ has been closed for a long time. At time $t = 0$, switch $S$ is opened. Which of the following expressions represents the magnitude of the current $I(t)$ through resistor $R_2$ as a function of time $t$ for $t \ge 0$? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Identify the target quantity and initial condition \[ I(0^+) = I_L(0^-) \]	The question asks for the current $I(t)$ through resistor $R_2$ for $t \ge 0$. Because current through an inductor cannot change instantaneously, the current in the discharging loop at $t = 0^+$ is equal to the steady-state current flowing through the inductor just before the switch opens at $t = 0^-$.
Calculate the steady-state inductor current prior to switch opening ($t = 0^-$) \[ R_p = \dfrac{R_2 R_3}{R_2 + R_3} = \dfrac{(2R)(R)}{2R + R} = \dfrac{2}{3}R \] \[ R_{\text{tot}} = R_1 + R_p = R + \dfrac{2}{3}R = \dfrac{5}{3}R \] \[ I_{\text{tot}} = \dfrac{V_0}{R_{\text{tot}}} = \dfrac{3V_0}{5R} \] \[ I_L(0^-) = I_{\text{tot}} \left( \dfrac{R_2}{R_2 + R_3} \right) = \left( \dfrac{3V_0}{5R} \right) \left( \dfrac{2R}{3R} \right) = \dfrac{2V_0}{5R} \]	In steady state, the inductor acts as a zero-resistance wire. We find the equivalent resistance of the parallel combination of $R_2$ and $R_3$, determine total battery current, and apply the current divider rule to find $I_L(0^-)$.
Formulate the loop differential equation for $t > 0$ \[ -L \dfrac{dI}{dt} – I R_3 – I R_2 = 0 \implies L \dfrac{dI}{dt} + 3R I = 0 \] \[ \tau = \dfrac{L}{R_{\text{loop}}} = \dfrac{L}{3R} \]	Opening switch $S$ disconnects the battery and $R_1$. The remaining active circuit is a single closed loop containing $R_2$, $R_3$, and $L$ in series with total resistance $R_{\text{loop}} = R_2 + R_3 = 3R$.
Solve the differential equation for $I(t)$ \[ \int_{I(0^+)}^{I(t)} \dfrac{dI’}{I’} = -\dfrac{3R}{L} \int_0^t dt’ \implies I(t) = \dfrac{2V_0}{5R} e^{-3Rt/L} \]	Integrating the differential equation using the initial condition $I(0^+) = \dfrac{2V_0}{5R}$ yields the exponential decay function for the current through the loop.

Why each choice is correct or incorrect:

(A) Incorrect. Uses the pre-switch steady-state current through $R_2$ ($V_0/5R$) as the initial current instead of enforcing current continuity through the inductor.

(B) Incorrect. Uses only $R_3 = R$ to compute the decay time constant instead of the combined resistance of the discharging loop ($R_2 + R_3 = 3R$).

(C) Correct. Accurately determines the initial inductor current via current division ($2V_0/5R$) and uses the full loop resistance ($3R$) to set the exponential decay rate.

(D) Incorrect. Uses the total battery current ($3V_0/5R$) as the initial inductor current, ignoring the parallel branch division between $R_2$ and $R_3$.

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A circuit consists of an ideal battery of constant emf $\varepsilon$, a resistor of resistance $R$, an inductor of inductance $L$, and an open switch connected in series. At time $t = 0$, the switch is closed. At the instant when the current in the circuit reaches half of its maximum steady-state value, what fraction of the total energy delivered by the battery up to that time is stored in the magnetic field of the inductor?

Step	Reasoning
Identify the target quantity and initial condition \[ I(0^+) = I_L(0^-) \]	The question asks for the current \(I(t)\) through resistor \(R_2\) for \(t \ge 0\). Because current through an inductor cannot change instantaneously, the current in the discharging loop at \(t = 0^+\) is equal to the steady-state current flowing through the inductor just before the switch opens at \(t = 0^-\).
Calculate the steady-state inductor current prior to switch opening (\(t = 0^-\)) \[ R_p = \dfrac{R_2 R_3}{R_2 + R_3} = \dfrac{(2R)(R)}{2R + R} = \dfrac{2}{3}R \] \[ R_{\text{tot}} = R_1 + R_p = R + \dfrac{2}{3}R = \dfrac{5}{3}R \] \[ I_{\text{tot}} = \dfrac{V_0}{R_{\text{tot}}} = \dfrac{3V_0}{5R} \] \[ I_L(0^-) = I_{\text{tot}} \left( \dfrac{R_2}{R_2 + R_3} \right) = \left( \dfrac{3V_0}{5R} \right) \left( \dfrac{2R}{3R} \right) = \dfrac{2V_0}{5R} \]	In steady state, the inductor acts as a zero-resistance wire. We find the equivalent resistance of the parallel combination of \(R_2\) and \(R_3\), determine total battery current, and apply the current divider rule to find \(I_L(0^-)\).
Formulate the loop differential equation for \(t > 0\) \[ -L \dfrac{dI}{dt} – I R_3 – I R_2 = 0 \implies L \dfrac{dI}{dt} + 3R I = 0 \] \[ \tau = \dfrac{L}{R_{\text{loop}}} = \dfrac{L}{3R} \]	Opening switch \(S\) disconnects the battery and \(R_1\). The remaining active circuit is a single closed loop containing \(R_2\), \(R_3\), and \(L\) in series with total resistance \(R_{\text{loop}} = R_2 + R_3 = 3R\).
Solve the differential equation for \(I(t)\) \[ \int_{I(0^+)}^{I(t)} \dfrac{dI’}{I’} = -\dfrac{3R}{L} \int_0^t dt’ \implies I(t) = \dfrac{2V_0}{5R} e^{-3Rt/L} \]	Integrating the differential equation using the initial condition \(I(0^+) = \dfrac{2V_0}{5R}\) yields the exponential decay function for the current through the loop.

Kinematics	Forces
\(\Delta x = v_i t + \frac{1}{2} at^2\)	\(F = ma\)
\(v = v_i + at\)	\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)	\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)	\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\)

Circular Motion	Energy
\(F_c = \frac{mv^2}{r}\)	\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)	\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)	\(KE_i + PE_i = KE_f + PE_f\)
	\(W = Fd \cos\theta\)

Momentum	Torque and Rotations
\(p = mv\)	\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)	\(I = \sum mr^2\)
\(p_i = p_f\)	\(L = I \cdot \omega\)

Simple Harmonic Motion	Fluids
\(F = -kx\)	\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)	\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)	\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)	\(F_b = \rho V g\)
\(a = -\omega^2 x\)	\(A_1v_1 = A_2v_2\)

Constant	Description
[katex]g[/katex]	Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]	Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]	Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]	Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]	Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]	Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]	Mass of the Sun

Variable	SI Unit
[katex]s[/katex] (Displacement)	[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)	[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)	[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)	[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)	[katex]\text{kilograms (kg)}[/katex]

Variable	Derived SI Unit
[katex]F[/katex] (Force)	[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)	[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)	[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)	[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)	[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)	[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)	[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)	[katex]\text{hertz (Hz)}[/katex]

Prefix	Symbol	Power of Ten
Pico-	p	[katex]10^{-12}[/katex]
Nano-	n	[katex]10^{-9}[/katex]
Micro-	µ	[katex]10^{-6}[/katex]
Milli-	m	[katex]10^{-3}[/katex]
Centi-	c	[katex]10^{-2}[/katex]
Deci-	d	[katex]10^{-1}[/katex]
(Base unit)	–	[katex]10^{0}[/katex]
Deca- or Deka-	da	[katex]10^{1}[/katex]
Hecto-	h	[katex]10^{2}[/katex]
Kilo-	k	[katex]10^{3}[/katex]
Mega-	M	[katex]10^{6}[/katex]
Giga-	G	[katex]10^{9}[/katex]
Tera-	T	[katex]10^{12}[/katex]

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AP Physics

Unit 1 - Vectors and Kinematics

MCQ

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Advanced

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