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#voltage lag

2 public questions tagged with this topic.

AC Fundamentals - RMS, Average and Peak Values

In an AC circuit with only an inductor, what is the phase relationship between the current and the voltage?

**Peak current** I_peak = V_peak/R for resistor, I_rms = V_rms/R, V_peak = √2 V_rms, for 200 V rms, V_peak=282.8 V, I_peak=282.8/80=3.535 A, rms I=200/80=2.5 A, average over complete cycle zero because positive and negative halves cancel. In a purely inductive AC circuit, the current lags the voltage by 90°. This is because the inductor opposes changes in current, causing the current to reach its peak after the voltage. 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 Current lags voltage by 90°, consistent wit

Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values

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AC Through Capacitor - Capacitive Reactance

In a purely capacitive AC circuit, the current leads the voltage by what phase angle?

**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. For a capacitor, i = i_m sin (ω t + (π/2)) , v = v_m sin ω t . Phase difference: Φ = (π/2) , so current leads voltage by 90° . 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,

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

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