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A 0.5 μF capacitor is connected across a 10 V DC battery. After a long time (steady state), what are the circuit current and the voltage across the capacitor?

Difficulty: Easy

Correct Answer: 0 A and 10 V

Explanation:


Introduction / Context:
Capacitors in DC circuits charge up to the supply voltage and then behave as open circuits. This question reinforces the steady-state behavior of a capacitor connected to an ideal battery through negligible resistance.


Given Data / Assumptions:

  • Capacitance C = 0.5 μF.
  • Supply: ideal 10 V DC battery.
  • Long time implies t ≫ 5RC; leakage and series resistance neglected.


Concept / Approach:
For a step DC input, the capacitor voltage rises exponentially: v_c(t) = 10(1 − e^{−t/RC}). The current is i(t) = C dv_c/dt = (10/R) e^{−t/RC}. As t → ∞, e^{−t/RC} → 0, so i(∞) → 0 A and v_c(∞) → 10 V. Hence, steady-state current is zero and the capacitor holds the full battery voltage.


Step-by-Step Solution:

Recognize steady state for DC: capacitor behaves as open circuit.Therefore i(∞) = 0 A.The capacitor voltage equals the source: v_c(∞) = 10 V.


Verification / Alternative check:

Energy stored W = 1/2 * C * V^2 = 0.5 * 0.5×10^{-6} * 10^2 J = 2.5×10^{-6} J, consistent with a charged, no-current state.


Why Other Options Are Wrong:

Options with nonzero steady DC current contradict the open-circuit nature of a charged capacitor on DC.Voltages other than 10 V violate KVL for an ideal battery in steady state.


Common Pitfalls:

Confusing instantaneous charging current with long-time behavior; ignoring that only transients carry current in capacitors under DC excitation.


Final Answer:

0 A and 10 V
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