The current and voltage in an a.c. circuit are given as
\( I = 50\sin(50t)\,\mathrm{mA} \)
\( V = 100\sin\left(50t + \displaystyle\frac{\pi}{3}\right)\,\mathrm{volt} \)
The power dissipated in the circuit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.25\,\mathrm{W} \)
The average power in an a.c. circuit is \( P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi \), where \( \phi \) is the phase difference between voltage and current. With peak values, \( V_{\mathrm{rms}}I_{\mathrm{rms}} = V_0I_0/2 \).
\( V_0 = 100\,\mathrm{V} \)
\( I_0 = 50\,\mathrm{mA} = 0.05\,\mathrm{A} \)
\( \phi = \pi/3 = 60^\circ \)
\( \displaystyle P = \frac{1}{2}\,V_0 I_0\cos\phi = \frac{1}{2} \times 100 \times 0.05 \times \cos60^\circ = 2.5 \times 0.5 = 1.25\,\mathrm{W} \)