Question

In a circuit, voltage and current are given by $V = 10 \sin(\omega t + 30^\circ)$ and $i = 10 \sin(\omega t - 30^\circ)$. The power consumed in the circuit is ____.

• A. 100 watts
• B. 50 watts
• C. 25 watts
• D. 12.5 watts

Hint:

Be careful with the phase! Always subtract the smaller angle from the larger one or follow the order $(V_{\text{angle}} - I_{\text{angle}})$. Here, $+30^\circ$ and $-30^\circ$ are $60^\circ$ apart on a phasor diagram.

Answer 1

The Correct Option is C

Step 1: Write the instantaneous power as the product of instantaneous voltage and current: \[ p(t) = 10\sin(\omega t + 30^\circ) \times 10\sin(\omega t - 30^\circ) \]

Step 2: Use the product-to-sum identity \( \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \), with \( A = \omega t + 30^\circ \) and \( B = \omega t - 30^\circ \), so \( A - B = 60^\circ \) and \( A + B = 2\omega t \): \[ p(t) = 100 \times \frac{1}{2}\left[\cos 60^\circ - \cos(2\omega t)\right] = 50\cos 60^\circ - 50\cos(2\omega t) \]

Step 3: Average this over one full cycle. The oscillating term \( \cos(2\omega t) \) averages to zero over a complete cycle, leaving only the constant term: \[ P_{avg} = 50\cos 60^\circ = 50 \times 0.5 = 25 \text{ W} \] \[ \boxed{P = 25 \text{ W}} \]