MHT CET · Physics · Alternating Current
An inductive coil has a resistance of \(100 \Omega\). When an a.c. signal of frequency \(1000 \mathrm{~Hz}\) is applied to the coil the voltage leads the current by \(45^{\circ}\). The inductance of the coil is
- A \(\frac{0.25}{2 \pi} \mathrm{H}\)
- B \(\frac{0.05}{\pi} \mathrm{H}\)
- C \(\frac{0.25}{\pi} \mathrm{H}\)
- D \(\frac{0.5}{\pi} \mathrm{H}\)
Answer & Solution
Correct Answer
(B) \(\frac{0.05}{\pi} \mathrm{H}\)
Step-by-step Solution
Detailed explanation
\(
\tan \phi=\tan 45^{\circ}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}
\)
or \(\mathrm{X}_{\mathrm{L}}=\mathrm{R}\)
\(
2 \pi \mathrm{fL}=\mathrm{R}
\)
Or \(L=\frac{\mathrm{R}}{2 \pi \mathrm{f}}=\frac{100}{2 \pi \times 1000}=\frac{0.05}{\pi} \mathrm{H}\)
\tan \phi=\tan 45^{\circ}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}
\)
or \(\mathrm{X}_{\mathrm{L}}=\mathrm{R}\)
\(
2 \pi \mathrm{fL}=\mathrm{R}
\)
Or \(L=\frac{\mathrm{R}}{2 \pi \mathrm{f}}=\frac{100}{2 \pi \times 1000}=\frac{0.05}{\pi} \mathrm{H}\)
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