Question

A voltage \( v = v_0 \sin(\omega t) \) applied to a circuit drives a current \( i = i_0 \sin(\omega t + \phi) \) in the circuit. The average power consumed in the circuit over a cycle is:

• A. Zero
• B. \( i_0 v_0 \cos \phi \)
• C. \( \frac{i_0 v_0}{2} \)
• D. \( \frac{i_0 v_0}{2} \cos \phi \)

Hint:

The average power in an AC circuit with sinusoidal voltage and current is given by \( P_{\text{avg}} = \frac{i_0 v_0}{2} \cos \phi \), where \( \phi \) is the phase difference.

Answer 1

The Correct Option is D

The average power \( P_{\text{avg}} \) consumed in an AC circuit is given by: \[ P_{\text{avg}} = \frac{1}{T} \int_0^T v(t) i(t) \, dt \] where \( T \) is the time period of the AC waveform. For the given voltage \( v = v_0 \sin(\omega t) \) and current \( i = i_0 \sin(\omega t + \phi) \), the instantaneous power is: \[ P(t) = v(t) i(t) = v_0 i_0 \sin(\omega t) \sin(\omega t + \phi) \] Using the trigonometric identity: \[ \sin(\omega t) \sin(\omega t + \phi) = \frac{1}{2} [\cos \phi - \cos(2 \omega t + \phi)] \] The average power over a cycle is the time average of this expression: \[ P_{\text{avg}} = \frac{v_0 i_0}{2} \cos \phi \] Thus, the correct answer is \( \frac{i_0 v_0}{2} \cos \phi \).