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#voltage peak

2 public questions tagged with this topic.

AC Fundamentals - RMS, Average and Peak Values

A \( 212.1 \, \text{V} \) (peak) AC source is connected to a \( 75 \, \Omega \) resistor. What is the average power cons

**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. RMS voltage: V = (v_m/√(2)) = (212.1/1.414) = 150 V . RMS current: I = (V/R) = (150/75) = 2 A . Average power: P = I² R = 2² × 75 = 300 W . 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 300 W, consistent

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

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AC Through Capacitor - Capacitive Reactance

In an AC circuit with only a capacitor, what is the nature of the current when the voltage is at its peak?

**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. In a purely capacitive circuit, the current leads the voltage by 90°. When the voltage is at its peak (maximum positive or negative), the current is zero because it reaches its peak 90° earlier and crosses zero at the voltage peak. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

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

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