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Question

Why does a purely capacitive AC circuit not dissipate power despite having current flow?

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Explanation

**Average power** in resistor P_avg = V_rms I_rms = V_rms²/R = I_rms² R, for 254.6 V peak, V_rms =180 V, R=90 Ω, P=180²/90=360 W, for 226.3 V peak, V_rms=160 V, R=80 Ω, P=320 W, illustrating rms use for power. In a purely capacitive circuit, the current leads the voltage by 90°. The instantaneous power oscillates, but the average power over a cycle is zero because the energy is stored during one half-cycle and returned during the other, with no net dissipation. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p =

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