Question:
A source voltage of 100 V dc is applied across a RLC series circuit. Under steady-state conditions, voltages across each element is
A source voltage of 100 V dc is applied across a RLC series circuit. Under steady-state conditions, voltages across each element is
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Under DC steady-state, inductor behaves as a short and capacitor as an open. Use this to quickly deduce element voltages.
Updated On: May 23, 2025
- \( V_R = 100V, V_L = 0, V_C = 0 \)
- \( V_R = 100V, V_L = -100V, V_C = 100V \)
- \( V_R = 0, V_L = 100V, V_C = -100V \)
- \( V_R = 0, V_L = 0, V_C = -100V \)
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The Correct Option is D
Solution and Explanation
In steady-state for a DC source:
- Inductor acts as a short circuit: \( V_L = 0 \)
- Capacitor acts as an open circuit: voltage across capacitor equals the source voltage with opposite polarity, i.e., \( V_C = -100V \)
- No current flows in the circuit, so \( V_R = 0 \) Thus: \[ V_R = 0, V_L = 0, V_C = -100V \]
- Inductor acts as a short circuit: \( V_L = 0 \)
- Capacitor acts as an open circuit: voltage across capacitor equals the source voltage with opposite polarity, i.e., \( V_C = -100V \)
- No current flows in the circuit, so \( V_R = 0 \) Thus: \[ V_R = 0, V_L = 0, V_C = -100V \]
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