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In a series R–C circuit excited by a step input of magnitude E at t = 0, what is the charge on the capacitor at the instant t = 0+?

Difficulty: Easy

0
C E
C E (1 - e^{-t/RC})
None of the above
C E e^{-t/RC}

Correct Answer: 0

Explanation:

Introduction / Context:
Immediately after a step voltage is applied to a series resistor–capacitor (R–C) circuit, the capacitor voltage and charge cannot change instantaneously because that would require an infinite current. This question probes transient behavior right at t = 0+.

Given Data / Assumptions:

Series R–C driven by a DC step of magnitude E at t = 0.
Capacitor is initially uncharged (q(0−) = 0).
Ideal components: no leakage or series inductance.

Concept / Approach:
For a step input E·u(t), the capacitor voltage is v_c(t) = E(1 − e^{−t/RC}) for t ≥ 0. The capacitor charge is q(t) = C v_c(t) = C E (1 − e^{−t/RC}). Evaluating at the instant after switching, t = 0+, gives q(0+) = C E (1 − 1) = 0. Physically, the capacitor behaves like a short circuit at t = 0+, so its voltage and charge are still zero while the current is maximum.

Step-by-Step Solution:

Write q(t) = C E (1 − e^{−t/RC}).Substitute t = 0+: q(0+) = C E (1 − 1) = 0.

Verification / Alternative check:

At t = 0+, v_c = 0 and i(0+) = E/R (finite), consistent with circuit laws.

Why Other Options Are Wrong:

C E is the final charge at steady state, not at t = 0+.C E (1 − e^{−t/RC}) is the time function; at 0+ it reduces to 0.Other forms conflict with the instantaneous behavior of capacitors.

Common Pitfalls:

Assuming the capacitor gets fully charged instantly; forgetting that charge/voltage cannot jump.

In a series R–C circuit excited by a step input of magnitude E at t = 0, what is the charge on the capacitor at the instant t = 0+?

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