Home / AP® Exam / AP® Physics C: Electricity and Magnetism / Circuits with Capacitors and Inductors (LC Circuits) MCQ
A circuit is constructed using a battery of emf \( \varepsilon \), a resistor of resistance \(R\), a capacitor of capacitance \(C\), an inductor of inductance \(L\), and three switches, as shown in the figure above. The three switches are labeled \(S_1\), \(S_2\), and \(S_3\), and they can be operated independently.
All switches are open, and there is no stored energy in the capacitor or the inductor. Switch \(S_3\) is closed. What is the current in the inductor after steady state has been reached?
Correct Answer: \( \boxed{\mathrm{D}} \)
Closing switch \(S_3\) connects the battery, resistor, and inductor in series, while the capacitor remains disconnected from the circuit.
After a long time (steady state), the current through the inductor becomes constant, so
\( \dfrac{dI}{dt}=0 \)
Therefore, the voltage across the inductor is
\( V_L=L\dfrac{dI}{dt}=0 \)
The inductor behaves like an ideal wire (short circuit), so the entire battery voltage appears across the resistor.
Applying Kirchhoff’s loop rule,
\( \varepsilon-IR=0 \)
which gives
\( I=\dfrac{\varepsilon}{R} \)
Thus, the steady-state current in the inductor is \( \boxed{\dfrac{\varepsilon}{R}} \).
Therefore, the correct answer is (D).
A circuit is constructed using a battery of emf \( \varepsilon \), a resistor of resistance \(R\), a capacitor of capacitance \(C\), an inductor of inductance \(L\), and three switches, as shown in the figure above. The three switches are labeled \(S_1\), \(S_2\), and \(S_3\), and they can be operated independently.
All switches are open, and there is no stored energy in the capacitor or the inductor. Switch \(S_1\) is closed. After the capacitor is fully charged, switch \(S_1\) is opened and switch \(S_2\) is closed. Which of the following expressions represents the maximum current in the LC circuit?
Correct Answer: \( \boxed{\mathrm{B}} \)
When switch \(S_1\) is closed, the capacitor charges completely through the resistor until its voltage equals the battery emf:
\( V_C=\varepsilon \)
The energy stored in the fully charged capacitor is
\( U_C=\dfrac{1}{2}C\varepsilon^2 \)
After \(S_1\) is opened and \(S_2\) is closed, the battery and resistor are disconnected, leaving an ideal LC circuit.
At the instant the current reaches its maximum value, all of the capacitor’s energy has been converted into magnetic energy stored in the inductor:
\( \dfrac{1}{2}C\varepsilon^2=\dfrac{1}{2}LI_{\max}^2 \)
Solving for the maximum current,
\( I_{\max}=\varepsilon\sqrt{\dfrac{C}{L}} \)
Thus, the maximum current in the LC circuit is \( \boxed{\varepsilon\sqrt{\dfrac{C}{L}}} \).
Therefore, the correct answer is (B).