Question

A 200-Ω resistor is connected across an ac voltage source V = (150 V) sin [2π(60 Hz)t]. What is the average power delivered to this circuit (in Watt)?

Answer 1

The formula for instantaneous power is:

P(t) = V(t)²/R = (150)² sin²[2pi(60)t]/R

If we average over a whole cycle we get:

Pavg = 1/(1/60) (150)² sin²[2pi(60)t]/R dt
Pavg = 1/(1/60 )(150)²/R sin²[2pi(60Hz)t] dt

but sin²(x) = 1/2 - cos(2x), so:

Pavg = 1/(1/60)(150)²/R 1/2 - cos[4pi(60)t] dt

= 1/(1/60)(150)²/R { 1/2 dt - cos[4pi(60Hz)t] dt }

= 1/(1/60)(150)²/R { 1/2 *t [at t = 1/60 s and 0 ] - (1/4pi(60))sin[4pi(60Hz)t] [at t = 1/60 s and 0 ] }

= (1/(1/60)(150)²/R { 1/120 s - (1/4pi(60Hz))(sin[4pi] - sin[0]) }

= 1/(1/60 s)(150)²/R { 1/120 s }

= 1/2(150)²/R

Pavg = 1/2*(150)²/200

Pavg = 56.25 W