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 consumed in a circuit is determined by first calculating the instantaneous power, which is the product of voltage \( v(t) \) and current \( i(t) \):

\[ p(t) = v(t) \cdot i(t) = (v_0 \sin(\omega t)) \cdot (i_0 \sin(\omega t + \phi)) \]

Applying the trigonometric identity \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \) with \( A = \omega t \) and \( B = \omega t + \phi \):

\[ \sin(\omega t) \sin(\omega t + \phi) = \frac{1}{2} [\cos(-\phi) - \cos(2\omega t + \phi)] \]

Since \( \cos(-\phi) = \cos(\phi) \), the instantaneous power becomes:

\[ p(t) = v_0 i_0 \frac{1}{2} [\cos(\phi) - \cos(2\omega t + \phi)] \]

To find the average power over a single period \( T = \frac{2\pi}{\omega} \), we integrate \( p(t) \) over \( T \) and divide by \( T \):

\[ P_{\text{avg}} = \frac{1}{T} \int_0^T v_0 i_0 \frac{1}{2} [\cos(\phi) - \cos(2\omega t + \phi)] \, dt \]

This integral can be split into two parts:

\[ P_{\text{avg}} = \frac{v_0 i_0}{2T} \left[\int_0^T \cos(\phi) \, dt - \int_0^T \cos(2\omega t + \phi) \, dt \right] \]

The first integral evaluates to:

\[ \int_0^T \cos(\phi) \, dt = \cos(\phi) \cdot T \]

The second integral, the integral of a cosine function over its period, is zero:

\[ \int_0^T \cos(2\omega t + \phi) \, dt = 0 \]

Substituting these results yields:

\[ P_{\text{avg}} = \frac{v_0 i_0}{2T} [\cos(\phi) \cdot T] = \frac{v_0 i_0}{2} \cos(\phi) \]

Therefore, the average power consumed in the circuit over a cycle is:

\[ \frac{i_0 v_0}{2} \cos \phi \]