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?
Electronics and Communication Engineering
Networks Analysis and Synthesis
Difficulty: Easy
Choose an option
-
A0.5 A and 0 V
-
B0 A and 10 V
-
C20 A and 5 V
-
D0.05 A and 5 V
-
E0 A and 0 V
Answer
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