AP Physics 2 · Question of the Day

AP Physics 2 Question of the Day

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Sunday, September 20, 2026

A capacitor is charging through a resistor from an ideal battery. Which statement best describes the resistor's voltage drop VRV_R as time increases?

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Question of the Day

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A capacitor is charging through a resistor from an ideal battery. Which statement best describes the resistor's voltage drop VRV_R as time increases?

  1. VRV_R starts near the battery voltage and decreases toward 0 V0\ \text{V}. (correct answer)
  2. VRV_R starts at 0 V0\ \text{V} and increases toward the battery voltage.
  3. VRV_R remains constant because the resistor's resistance is constant.
  4. VRV_R becomes greater than the battery voltage briefly, then returns.

Explanation: This question tests understanding of resistor-capacitor (RC) circuits. When charging begins, the uncharged capacitor acts like a short circuit with zero voltage, so the entire battery voltage initially appears across the resistor, creating maximum current (I₀ = V_battery/R). As the capacitor charges and its voltage increases, the voltage across the resistor must decrease to maintain Kirchhoff's voltage law (V_battery = V_R + V_C), causing the current and resistor voltage to decay exponentially toward zero. At long times, the capacitor reaches battery voltage, current stops, and the resistor voltage becomes zero. Choice C incorrectly assumes constant resistor voltage, representing the misconception that resistance alone determines voltage drop—in reality, V_R = IR depends on the changing current. To solve charging circuits, apply Kirchhoff's voltage law at any instant to relate component voltages.