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)?
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
A 200-Ω resistor is connected across an ac voltage source V = (150 V) sin [2π(60...
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