Current and voltage in an a.c. circuit are given as
\( I = 10\sin(50t - \pi/2)\,\mathrm{A} \)
\( V = 200\sin(50t)\,\mathrm{V} \) (t represents time)
The power dissipated in the circuit is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 0\,\mathrm{W} \)
The average power in an a.c. circuit depends on the phase difference \( \phi \) between voltage and current through the power factor \( \cos\phi \).
\( \displaystyle P = V_{\mathrm{rms}}\,I_{\mathrm{rms}}\cos\phi \)
Here the current lags the voltage by \( \phi = \pi/2 \) (a purely reactive circuit).
\( \displaystyle \cos\phi = \cos\frac{\pi}{2} = 0 \;\Rightarrow\; P = 0\,\mathrm{W} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Power in AC Circuit: The Power Factor – “… So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle \( \phi \) between them. The quantity \( \cos\phi \) is called the power factor. … Case (ii) Purely inductive or capacitive circuit: If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is \( \pi/2 \). Therefore, \( \cos\phi = 0 \), and no power is dissipated even though a current is flowing in the circuit.”