In an ac circuit, the applied voltage is \( V=220\sin(100t) \) volt. If the impedance is \( 110\,\Omega \) and the phase difference between the current and the voltage is \( 60^\circ \), the power consumed is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 110\,\mathrm{W} \)
The average power in an ac circuit is \( P=\frac{1}{2}v_mi_m\cos\phi \), where \( v_m \) and \( i_m \) are the peak values and \( \cos\phi \) is the power factor. Comparing with \( v=v_m\sin\omega t \), the peak voltage is \( v_m=220\,\mathrm{V} \).
The peak current is:
\( \begin{aligned} i_m &= \frac{v_m}{Z} \\ &= \frac{220\,\mathrm{V}}{110\,\Omega} \\ &= 2\,\mathrm{A} \end{aligned} \)
The power consumed is:
\( \begin{aligned} P &= \frac{1}{2}v_mi_m\cos\phi \\ &= \frac{1}{2}\times220\,\mathrm{V}\times2\,\mathrm{A}\times\cos60^\circ \\ &= 220\,\mathrm{W}\times0.5 \\ &= 110\,\mathrm{W} \end{aligned} \)