Home / AP® Exam / AP® Physics C: Electricity and Magnetism / Resistor-Capacitor (RC) Circuits MCQ
What is the value of the following product?
\(20~\mu\mathrm{F}\times500~\Omega\)
Correct Answer: \( \boxed{\mathrm{D}} \)
The time constant of an RC circuit is
\( \tau=RC \)
Substitute the given values:
\( \tau=(20\times10^{-6}~\mathrm{F})(500~\Omega)=0.01~\mathrm{s} \)
Alternatively, verify using units:
\( \mathrm{F}\cdot\Omega=\dfrac{\mathrm{C}}{\mathrm{V}}\cdot\dfrac{\mathrm{V}}{\mathrm{A}}=\dfrac{\mathrm{C}}{\mathrm{A}}=\mathrm{s} \)
Thus, the product has the dimensions of time and equals
\( \boxed{0.01~\mathrm{s}} \)
Hence, the correct answer is (D).
An initially uncharged \(3.0\,\mu\mathrm{F}\) capacitor is placed in a circuit with an ideal \(30\,\mathrm{V}\) battery, two resistors, and an open switch \(S\), as shown in the figure above. The switch is then closed.
What is the current in the \(10\,\Omega\) resistor immediately after the switch is closed?
Correct Answer: \( \boxed{\mathrm{D}} \)
Immediately after the switch is closed, the capacitor is initially uncharged, so its voltage is zero:
\(V_C(0)=0\)
Therefore, the capacitor behaves like a short circuit at the instant the switch is closed.
The capacitor shorts out the \(20\,\Omega\) resistor, so no current initially flows through the \(20\,\Omega\) branch.
The circuit effectively contains only the \(10\,\Omega\) resistor connected to the \(30\,\mathrm{V}\) battery.
Applying Ohm’s law,
\(I=\dfrac{V}{R}=\dfrac{30\,\mathrm{V}}{10\,\Omega}=3.0\,\mathrm{A}\)
Thus, the initial current through the \(10\,\Omega\) resistor is
\(\boxed{3.0\,\mathrm{A}}\).
Therefore, the correct answer is (D).
A capacitor of capacitance \(C\) is connected in series with a resistor of resistance \(R\) and a battery of emf \(\varepsilon\). The graph above shows the charge \(q\) on the capacitor approaching a value \(q_{\mathrm{max}}\) with increasing time \(t\).
What is \(q_{\mathrm{max}}\)?
Correct Answer: \( \boxed{\mathrm{E}} \)
As time increases, the capacitor becomes fully charged and the current in the circuit approaches zero. At this steady state, the voltage across the capacitor equals the battery emf:
\(V_C=\varepsilon\)
Using the definition of capacitance,
\(Q=CV\),
the maximum charge stored on the capacitor is
\(q_{\mathrm{max}}=C\varepsilon\).
The resistance \(R\) determines only how quickly the capacitor charges through the time constant
\(\tau=RC\),
but it does not affect the final charge on the capacitor.
Therefore, the correct answer is (E).