MHT CET · Physics · Alternating Current
When a coil is connected to a d.c. source of e.m.f. 12 volt, then the current of 4 A flows in it. If the same coil is connected to a 12 volt, 50 Hz a.c. source, then the current flowing in it is 2.4 A . Then self-inductance of the coil will be
- A 48 H
- B 12 H
- C \(\frac{4}{\pi} \times 10^{-2} \mathrm{H}\)
- D \(\frac{8}{\pi} \times 10^{-2} \mathrm{H}\)
Answer & Solution
Correct Answer
(C) \(\frac{4}{\pi} \times 10^{-2} \mathrm{H}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{12}{4}=3 \Omega(\text { d.c. source }) \\ & \mathrm{Z}=\frac{12}{2.4}=5 \Omega(\text { a.c. source }) \\ & \mathrm{Z}^2=\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2 \\ \therefore \quad & \mathrm{X}_{\mathrm{L}}^2=25-9=16 \\ \therefore \quad & \mathrm{X}_{\mathrm{L}}=4 \\ & \text { Also, } X_L=\omega \mathrm{L}=2 \pi \mathrm{fL} \\ \therefore \quad & \mathrm{L}=\frac{X_L}{2 \pi \mathrm{f}}=\frac{4}{2 \pi \times 50}=\frac{4}{100 \pi}=\frac{4}{\pi} \times 10^{-2} \mathrm{H}\end{array}\)
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