Question

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

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

Hint:

The average power in an AC circuit is the product of the RMS values of voltage and current and the cosine of the phase difference.

Answer 1

The Correct Option is B

The circuit's voltage is \( v = v_0 \sin(\omega t) \) and its current is \( i = i_0 \sin(\omega t + \varphi) \). The objective is to determine the average power consumed over one cycle. The instantaneous power \( p(t) \) is computed as:

\[ p(t) = v(t) \times i(t) = v_0 \sin(\omega t) \times i_0 \sin(\omega t + \varphi) \]

Applying the product-to-sum trigonometric identity, \(\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]\), yields:

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

To ascertain the average power over a single cycle (with period \( T = \frac{2\pi}{\omega} \)), we average \( p(t) \):

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

The average value of the \(\cos(2\omega t + \varphi)\) term across a cycle is zero, as it completes an integer number of cycles within the period. Consequently,

\[ \text{Average Power} = \frac{1}{T} \int_0^T \frac{1}{2} i_0 v_0 \cos(\varphi) \, dt = \frac{1}{2} i_0 v_0 \cos(\varphi) \]

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

\( \frac{1}{2} i_0 v_0 \cos \varphi \)