Difficulty: Easy
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:
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.Final Answer:
0
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