When will capacitor fully charged?

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When the supply voltage is equal to the capacitor voltage Explanation: A capacitor is considered fully charged when the potential difference across its plates becomes equal to the applied DC supply voltage. At this point, the charging current drops to zero, and the capacitor acts as an open circuit in a steady-state DC condition. Working Principle: As the capacitor charges, it accumulates charge $q = Cv$ on its plates. This creates a back EMF (or plate voltage) that opposes the supply voltage. As the voltage across the capacitor $v_c(t)$ increases, the charging current $i(t) = \frac{V_s - v_c(t)}{R}$ decreases exponentially. The process reaches steady state when $v_c(t) = V_s$, resulting in zero potential difference across the charging resistor. Key Formulas: $v_c(t) = V_s(1 - e^{-\frac{t}{RC}})$ — Instantaneous voltage across capacitor; $i(t) = \frac{V_s}{R} e^{-\frac{t}{RC}}$ — Charging current at time t Key Points: Steady state is theoretically achieved at $t = 5 au$, where $ au = RC$ is the time constant.; At steady state, the capacitor stores energy $E = \frac{1}{2}CV^2$.; Once charged, the capacitor behaves like an open switch in a DC circuit. Advantages: Energy storage capability; Filters voltage ripples in DC supplies Disadvantages: Cannot store energy indefinitely due to leakage resistance; Finite time required for charging Applications: Timing circuits (timers); Smoothing filters in power supplies Additional Info: Option B is incorrect because $1/\sqrt{2}$ is related to RMS values in AC circuits or half-power frequencies, not DC charging states.. Option D is incorrect as the transient charging process is finite for practical circuits. Answer: C is correct — A capacitor reaches full charge when the potential difference across it matches the source voltage, terminating the charging current. Exam Tip: Remember that in DC circuits, a capacitor acts as a short circuit at $t=0$ and as an open circuit at $t=\infty$.

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