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

To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage \( v(t) \) and the current \( i(t) \):

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

Using the trigonometric identity for the product of sines:

\[ \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \]

Substitute \( A = \omega t \) and \( B = \omega t + \phi \):

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

\(\cos(-\phi) = \cos(\phi)\), thus:

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

To find the average power over a cycle, integrate this expression over one period \( T = \frac{2\pi}{\omega} \), then 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 separates into two integrals:

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

The first term simplifies to:

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

And for the second term, the integral of cosine over a period is zero:

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

Therefore:

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

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

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