Question

A load consisting of a 480 resistor in parallel with a (5/9) μF capacitor is connected across the terminals of a sinusoidal voltage source vg, where vg = 100 cos5000t V. What is the peak value of the instantaneous power delivered by the source?

Answer 1

The instantaneous power delivered by a sinusoidal voltage source is given by:Instantaneous power delivered P(t) = Vg2 / R × cos2(ωt - Φ) / (1 + ω2R2C2)Where,Vg = Peak voltage of sinusoidal voltage sourceR = Value of resistanceC = Value of capacitanceω = Angular frequency of sinusoidal voltage source, given as 2πf where f is the frequency of the sourceΦ = Phase angle between current and voltageTherefore, for the given circuit, we have;R = 480 ΩC = 5/9 μF = 5 × 10⁻⁹ FVg = 100 Vω = 2πf = 2π × 5000 rad/s = 10⁵π rad/sΦ = 0 (since the voltage and current are in phase for a purely resistive circuit)Substituting the given values, we get;Instantaneous power delivered P(t) = (100/√2)² / 480 × cos²(10⁵πt) / (1 + 480² × 5² × 10⁻¹⁸)On solving the above expression, we get;P(t) = 106.25 cos²(10⁵πt) WThus, the peak value of the instantaneous power delivered by the source is 106.25 W.Answer: 106.25 W.