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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 ?
- A \(\frac{\pi}{6} \mathrm{rad}\)
- B \(\frac{\pi}{3} \mathrm{rad}\)
- C \(\frac{\pi}{4} \mathrm{rad}\)
- D \(\frac{\pi}{2} \mathrm{rad}\)
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)
\(
\begin{aligned}
\mathrm{P} & =\mathrm{P}_{\max } \cos \phi \\
\therefore \mathrm{P}_{\max } \times \frac{50}{100} & =\mathrm{P}_{\max } \cos \phi \\
\therefore \quad \cos \phi & =\frac{1}{2} \\
\phi & =\frac{\pi}{3} \mathrm{rad}
\end{aligned}
\)
\(\mathrm{P}=\mathrm{VI} \cos \phi\) \(\qquad\)
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)
\(
\begin{aligned}
\mathrm{P} & =\mathrm{P}_{\max } \cos \phi \\
\therefore \mathrm{P}_{\max } \times \frac{50}{100} & =\mathrm{P}_{\max } \cos \phi \\
\therefore \quad \cos \phi & =\frac{1}{2} \\
\phi & =\frac{\pi}{3} \mathrm{rad}
\end{aligned}
\)
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