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 voltage applied to the circuit is given by \( v = v_0 \sin(\omega t) \) and the current in the circuit is given by \( i = i_0 \sin(\omega t + \varphi) \). The task is to find the average power consumed over a cycle. The instantaneous power \( p(t) \) is calculated as:

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

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

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

To find the average power over one cycle (period \( T = \frac{2\pi}{\omega} \)), we take the average of \( p(t) \):

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

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

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

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

\( i_0 v_0 \cos \varphi \)