Practice question
Question
A \( 40 \, \Omega \) resistor is connected to a \( 120 \, \text{V} \) (rms) AC source. What is the rms
current?
Explanation
**AC through capacitor** voltage lags current by 90°, Q= C V, I= dQ/dt = C dV/dt, V(t)=V_peak sin ωt, I(t)=I_peak sin(ωt+90°), average power zero because energy stored in electric field ½ C V² returned each cycle, capacitor blocks DC but passes AC, X_C decreases with f. RMS current: I = (V/R) . Given: V = 120 V , R = 40 Ω . I = (120/40) = 3 A . 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 = N_s/N_p, calculation gives 3 A, consistent with phasor analysis and resonance condition X_L = X_C.
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