Q123 / 180 · AIIMS 2008 · Physics
When \(100 \mathrm{~V}\) d.c. is applied across a coil, a current of 1 A flows through it. When \(100 \mathrm{~V}\) a.c. of \(50 \mathrm{~Hz}\) is applied to the same coil only 0.5 A flows. The inductance of the coil is
- Correct answerA\(0.55 \mathrm{~H}\)
- B\(5.5 \mathrm{~mH}\)
- C\(0.55 \mathrm{~mH}\)
- D\(55 \mathrm{~mH}\).
Solution
For dc, $V=I R$ ( $R$ is resistance of inductor)
$100=1 \times R \Rightarrow R=100 \Omega$.
For ac, $Z=\sqrt{R^2+X_L^2}$
$\begin{aligned} & \because V=I Z \\ & \Rightarrow 100=0.5 \sqrt{R^2+X_L^2} \\ & \Rightarrow 100=0.5 \sqrt{(100)^2+X_L^2}\end{aligned}$
$\begin{aligned} & \Rightarrow 100=\frac{5}{10} \sqrt{(100)^2+X_L^2} \\ & X_L^2=(200)^2-(100)^2=300 \times 100 \\ & \Rightarrow X_L=\sqrt{3 \times(100)^2}=100 \sqrt{3} \\ & \Rightarrow \omega L=100 \sqrt{3} \Rightarrow 2 \pi v L=100 \sqrt{3} \\ & \Rightarrow 2 \times \pi \times 50 \times L=100 \sqrt{3} \\ & \Rightarrow 100 \times \pi L=100 \sqrt{3} \\ & \therefore L=\frac{\sqrt{3}}{\pi}=\frac{\sqrt{3}}{3.14} \Rightarrow L=0.55 \mathrm{H} .\end{aligned}$
Hence, the correct answer is option: 'A'.