Question

The instantaneous values of alternating current and voltages in a circuit given as
$$ i = \frac{1}{\sqrt{2}} \sin(100\pi t) \text{ amp} $$ $$ e = \frac{1}{\sqrt{2}} \sin(100\pi t + \pi/3) \text{ volt} $$ The average power (in watts) consumed in the circuit is

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{\sqrt{3}}{2} \)
• D. \( \frac{1}{2} \)

Hint:

When calculating average power in AC circuits, use the formula \( P = \frac{1}{2} V_{\text{rms}} I_{\text{rms}} \), where \( V_{\text{rms}} \) and \( I_{\text{rms}} \) are the RMS values of voltage and current.

Answer 1

The Correct Option is D

The instantaneous power in an AC circuit is given by the product of the instantaneous current and voltage: \[ p(t) = e(t) \cdot i(t) \] Substitute the given expressions for \( e(t) \) and \( i(t) \): \[ p(t) = \frac{1}{\sqrt{2}} \sin(100\pi t + \frac{\pi}{3}) \cdot \frac{1}{\sqrt{2}} \sin(100\pi t) \] Using the trigonometric identity for the product of sines: \[ \sin A \cdot \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \] This gives: \[ p(t) = \frac{1}{2} \left[ \cos \left(\frac{\pi}{3} \right) - \cos \left(2 \cdot 100 \pi t + \frac{\pi}{3} \right) \right] \] The average power over one period of the wave is the time average of \( p(t) \), and the second cosine term averages to zero over one full cycle. The result is: \[ P_{\text{avg}} = \frac{1}{2} \cdot \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2} \]
Thus, the average power consumed in the circuit is \( \frac{1}{2} \).