Log inSign up
AP Physics C Electricity and Magnetism · Question of the Day

AP Physics C Electricity and Magnetism Question of the Day

A fresh daily question to build accuracy, reinforce recall, and turn practice into a steady habit.
Monday, September 21, 2026

A capacitor of capacitance CC holds an initial charge Q0Q_0. At time t=0t=0, a switch is closed to connect the capacitor in series with a resistor of resistance RR. Which expression represents the current I(t)I(t) in the resistor as a function of time, where positive current is defined as flowing away from the initially positive plate?

Keep practicing AP Physics C Electricity and Magnetism

Question of the Day

Answer today's AP Physics C Electricity and Magnetism question, reveal the full explanation, then keep the streak going with a new question every day.

A capacitor of capacitance CC holds an initial charge Q0Q_0. At time t=0t=0, a switch is closed to connect the capacitor in series with a resistor of resistance RR. Which expression represents the current I(t)I(t) in the resistor as a function of time, where positive current is defined as flowing away from the initially positive plate?

  1. I(t)=Q0RCet/RCI(t) = \frac{Q_0}{RC} e^{t/RC}
  2. I(t)=−Q0RC(1−e−t/RC)I(t) = -\frac{Q_0}{RC} (1 - e^{-t/RC})
  3. I(t)=Q0Ce−t/RCI(t) = \frac{Q_0}{C} e^{-t/RC}
  4. I(t)=Q0RCe−t/RCI(t) = \frac{Q_0}{RC} e^{-t/RC} (correct answer)

Explanation: During discharge, the charge on the capacitor decreases exponentially: Q(t)=Q0e−t/RCQ(t) = Q_0 e^{-t/RC}. The current I(t)I(t) is the rate at which charge flows through the resistor. Since the charge on the capacitor is decreasing, the current flowing out of the positive plate is I(t)=−dQ/dtI(t) = -dQ/dt. Differentiating Q(t)Q(t) with respect to time gives dQ/dt=Q0(−1RC)e−t/RCdQ/dt = Q_0(-\frac{1}{RC})e^{-t/RC}. Therefore, I(t)=−(−Q0RCe−t/RC)=Q0RCe−t/RCI(t) = -(-\frac{Q_0}{RC}e^{-t/RC}) = \frac{Q_0}{RC}e^{-t/RC}.