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:

In AC circuits, the power factor \( \cos \phi \) determines the actual power consumption. It accounts for phase differences between voltage and current.

Answer 1

The Correct Option is D

This problem concerns the calculation of average power dissipated in an AC circuit over a full cycle. In an AC circuit, voltage and current are defined as follows:

Voltage: \( v = v_0 \sin \omega t \)

Current: \( i = i_0 \sin (\omega t + \phi) \)

The average power \( P \) is determined by the integral:

\( P = \frac{1}{T} \int_0^T v \cdot i \, dt \)

Where \( T \) represents the waveform's period. Substituting the voltage and current expressions yields:

\( P = \frac{1}{T} \int_0^T v_0 \sin \omega t \cdot i_0 \sin (\omega t + \phi) \, dt \)

\( = \frac{v_0 i_0}{T} \int_0^T \sin \omega t \cdot \sin (\omega t + \phi) \, dt \)

Applying the trigonometric identity \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \), we get:

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

Substituting this back into the power equation:

\( P = \frac{v_0 i_0}{2T} \left( \int_0^T \cos \phi \, dt - \int_0^T \cos(2\omega t + \phi) \, dt \right) \)

Given that \( \cos(\phi) \) is a constant with respect to \( t \) and the integral of a cosine function over a complete period is zero:

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

\( \int_0^T \cos \phi \, dt = T \cos \phi \)

Therefore, the average power is:

\( P = \frac{v_0 i_0}{2} \cos \phi \)

Consequently, the average power consumed in the circuit over a cycle is \( \frac{i_0 v_0}{2} \cos \phi \).