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A capacitor in a DC circuit, in STEADY STATE, acts as
Aa short circuit
Ba resistor of value $1/C$
Can open circuit
Dan ideal voltage source of $V = 0$
Answer & Solution
Correct answer: C. an open circuit
1. CAPACITOR current: $i_C = C\,dV/dt$.
2. In DC STEADY STATE: $dV/dt = 0$ (voltage is constant), so $i_C = 0$.
3. ZERO current means infinite impedance → OPEN CIRCUIT.
4. CONTRAST: an INDUCTOR in DC steady state acts as a SHORT (voltage $L\,di/dt = 0$ when current is constant).
5. Option A is the inductor's DC steady state. Options B, D are wrong.
_Source: Tony Kuphaldt, "Lessons in Electric Circuits — DC", Vol I, Ch 13 (Capacitors in DC)._
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