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GUJCET · Physics · Alternating Current
A circuit is made of \(1 \Omega\) resistance and 2.5 mH inductance in series. If alternating source of 200 V and 50 Hz is connected with circuit, then the phase difference between current and voltage is ____________ .
- A \(\tan ^{-1} \pi\)
- B \(\tan ^{-1}\left(\frac{\pi}{2}\right)\)
- C \(\tan ^{-1}\left(\frac{\pi}{4}\right)\)
- D \(\tan ^{-1}\left(\frac{\pi}{3}\right)\)
Answer & Solution
Correct Answer
(C) \(\tan ^{-1}\left(\frac{\pi}{4}\right)\)
Step-by-step Solution
Detailed explanation
(C) \(\tan ^{-1}\left(\frac{\pi}{4}\right)\)
\(
\begin{aligned}
\tan \phi & =\frac{\omega \mathrm{L}}{\mathrm{R}} \\
\tan \phi & =\frac{2 \pi \nu \mathrm{~L}}{\mathrm{R}} \\
\tan \phi & =\frac{2 \pi \times 50 \times 2.5 \times 10^{-3}}{1} \\
\therefore \tan \phi & =\frac{\pi}{4} \\
\phi & =\tan ^{-1}\left(\frac{\pi}{4}\right)
\end{aligned}
\)
\(
\begin{aligned}
\tan \phi & =\frac{\omega \mathrm{L}}{\mathrm{R}} \\
\tan \phi & =\frac{2 \pi \nu \mathrm{~L}}{\mathrm{R}} \\
\tan \phi & =\frac{2 \pi \times 50 \times 2.5 \times 10^{-3}}{1} \\
\therefore \tan \phi & =\frac{\pi}{4} \\
\phi & =\tan ^{-1}\left(\frac{\pi}{4}\right)
\end{aligned}
\)
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