MHT CET · Physics · Alternating Current
An inductance coil has a resistance of \(80 \Omega\). When on AC signal of frequency 480 Hz is applied to the coil, the voltage leads the current by \(45^{\circ}\). The inductance of the coil in henry is \(\left[\sin 45^{\circ}=\cos 45^{\circ}=1 / \sqrt{2}\right]\)
- A \(\frac{1}{24 \pi}\)
- B \(\frac{\pi}{20}\)
- C \(\frac{\pi}{40}\)
- D \(\frac{1}{12 \pi}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{12 \pi}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \tan \phi=\frac{X_L}{R} \\ & \tan 45=1=\frac{2 \pi \mathrm{fL}}{\mathrm{R}} \\ \therefore \quad & L=\frac{R}{2 \pi \mathrm{f}}=\frac{80}{2 \pi \times 480}=\frac{1}{12 \pi}\end{array}\)
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