In an a.c. circuit, voltage and current are given by : V=100sin(100t)V and I=100sin(100t+π3)mA respectively.The average power dissipated in one cycle is :
Question
In an a.c. circuit, voltage and current are given by :
V = 100sin(100t)V
I = 100sin(100t + \frac{\pi}{3})mA
respectively. The average power dissipated in one cycle is :
Solution
The average power dissipated in an AC circuit is given by the formula Pavg = VIcos(Φ), where V is the peak voltage, I is the peak current, and Φ is the phase difference between the voltage and the current.
In this case, the peak voltage V is 100V, the peak current I is 100mA = 0.1A, and the phase difference Φ is π/3 rad.
Substituting these values into the formula gives:
Pavg = (100V)(0.1A)cos(π/3) = 10Wcos(π/3) = 5W.
So, the average power dissipated in one cycle is 5 Watts.
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