ExamBro
ExamBro
GUJCET · Physics · Alternating Current

A lamp consumes only \(50 \%\) of maximum power in an AC circuit. What is the phase difference between the applied voltage and the circuit current ?

  1. A \(\frac{\pi}{6} \mathrm{rad}\)
  2. B \(\frac{\pi}{3} \mathrm{rad}\)
  3. C \(\frac{\pi}{4} \mathrm{rad}\)
  4. D \(\frac{\pi}{2} \mathrm{rad}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\pi}{3} \mathrm{rad}\)

Step-by-step Solution

Detailed explanation

(B) \(\frac{\pi}{3} \mathrm{rad}\)
\(\mathrm{P}=\mathrm{VI} \cos \phi\) \(\qquad\)
If \(\cos \phi=1\) then \(\mathrm{P}=\mathrm{P}_{\text {max }}\)
\(
\begin{equation*}
\therefore \quad \mathrm{P}_{\max }=\mathrm{VI} \qquad\ldots(2)
\end{equation*}
\)
from equation (1) & (2)
\(
\begin{aligned}
\mathrm{P} & =\mathrm{P}_{\max } \cos \phi \\
\therefore \mathrm{P}_{\max } \times \frac{50}{100} & =\mathrm{P}_{\max } \cos \phi \\
\therefore \quad \cos \phi & =\frac{1}{2} \\
\phi & =\frac{\pi}{3} \mathrm{rad}
\end{aligned}
\)