A) 0.1 mH
B) 1 mH
C) 0.1 H
D) 1.1 H
Correct Answer: D
Solution :
Current through the bulb\[(I)=\frac{P}{V}\] \[=\frac{50}{100}=0.5\,A\] Impedance (Z) of the circuit \[=\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}\] ?(i) Inductive reactance\[({{X}_{L}})\] \[=\omega L=2\pi f.\,L=100\,\pi \,L\] Resistance of bulb\[(R)=\frac{{{V}^{2}}}{P}=\frac{{{(100)}^{2}}}{50}=200\,\Omega \] Impedance of the circuit \[(Z)=\frac{200}{0.5}=400\,\Omega \] Now, \[X_{L}^{2}={{Z}^{2}}-{{R}^{2}}\] \[={{(400)}^{2}}-{{(200)}^{2}}=12\times {{10}^{4}}\] \[{{(100\pi L)}^{2}}=12\times {{10}^{4}}\] \[L=\frac{\sqrt{12\times {{10}^{4}}}}{100\,\pi }=\frac{2\sqrt{3}}{\pi }=1.1\,H.\]
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