What is the initial current while charging a capacitor?

Question

Examoogle · Accepted Answer

High

Explanation: At the exact instant a capacitor is connected to a DC voltage source, it acts as a short circuit because the potential difference across its plates is zero. Consequently, the initial charging current is at its maximum possible value, determined solely by the source voltage and the series resistance of the circuit.

Working Principle: According to the fundamental equation of a capacitor, $i(t) = C \frac{dv}{dt}$. At $t=0$, the voltage across the capacitor $v_c(0) = 0$. Since the voltage cannot change instantaneously, the derivative $\frac{dv}{dt}$ is at its maximum, causing the current $i(0) = \frac{V_s}{R}$ to be at its peak. As the capacitor charges, $v_c$ increases, the voltage drop across the resistor decreases, and the charging current decays exponentially toward zero.

Key Formulas: $i(t) = \frac{V_s}{R} e^{-\frac{t}{RC}}$ — Time-domain expression for charging current; $I_{initial} = \frac{V_s}{R}$ — Current at $t=0$

Key Points: A capacitor blocks DC current at steady state ($t = \infty$).; A capacitor acts as a short circuit at the instant of switching ($t = 0$).; The initial current is limited only by the total series resistance in the circuit.; The time constant $	au = RC$ determines the rate at which the current decays.

Advantages: Fast initial energy storage response; Enables simple RC timing circuits

Disadvantages: High inrush current can stress power supplies; Requires current limiting to prevent component damage

Applications: Power supply filtering; Timing and oscillation circuits; Coupling and decoupling circuits

Additional Info: The duration of the high current phase is governed by the time constant $	au = RC$.. Option B (Low) is incorrect as the current is maximum at the start.. Option C (0) represents the steady-state condition ($t 	o \infty$), not the initial state.

Answer: A is correct — The initial charging current is high because the capacitor behaves like a short circuit when it is completely uncharged.

Exam Tip: Always remember: Capacitors resist changes in voltage ($V_c(0^+) = V_c(0^-)$), whereas inductors resist changes in current ($I_L(0^+) = I_L(0^-)$).

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