MHT CET · Physics · Electromagnetic Induction
A coil having an inductance of \(\frac{1}{\pi} \mathrm{H}\) is connected in series with a resistance of \(300 \Omega\). If \(20 \mathrm{~V}\) from a \(200 \mathrm{~Hz}\) source are impressed across the combination, the value of the phase angle between the voltage and the current is
- A \(\tan ^{-1}\left(\frac{5}{4}\right)\)
- B \(\tan ^{-1}\left(\frac{4}{5}\right)\)
- C \(\tan ^{-1}\left(\frac{3}{4}\right)\)
- D \(\tan ^{-1}\left(\frac{4}{3}\right)\)
Answer & Solution
Correct Answer
(D) \(\tan ^{-1}\left(\frac{4}{3}\right)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{X}_{\mathrm{L}}=\mathrm{L} \omega=\mathrm{L} \times 2 \pi \mathrm{f} \\
\therefore & \mathrm{X}_{\mathrm{L}}=\frac{1}{\pi} \times 2 \pi \times 200 \\
\therefore & \mathrm{X}_{\mathrm{L}}=400 \Omega
\end{aligned}\)
Now, the phase angle' between voltage and current is given by, \(\tan \phi \stackrel{X_L}{R}=\frac{400}{300}\)
\(\therefore \quad \phi=\tan ^{-1}\left(\frac{4}{3}\right)\)
& \mathrm{X}_{\mathrm{L}}=\mathrm{L} \omega=\mathrm{L} \times 2 \pi \mathrm{f} \\
\therefore & \mathrm{X}_{\mathrm{L}}=\frac{1}{\pi} \times 2 \pi \times 200 \\
\therefore & \mathrm{X}_{\mathrm{L}}=400 \Omega
\end{aligned}\)
Now, the phase angle' between voltage and current is given by, \(\tan \phi \stackrel{X_L}{R}=\frac{400}{300}\)
\(\therefore \quad \phi=\tan ^{-1}\left(\frac{4}{3}\right)\)
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