A) \[0.1\,mH\]
B) \[1\,mH\]
C) \[0.1\,H\]
D) \[1.1\,H\]
Correct Answer: D
Solution :
Resistance of bulb, \[R=\frac{{{V}^{2}}}{P}\,=\frac{{{(100)}^{2}}}{50}\,-200\Omega \] Current through bulb, \[(I)=\frac{V}{R}\,=\frac{100}{200}\,=0.5\,A\] In a circuit containing inductive reactance \[({{X}_{L}})\] and resistance (R), Impedance (Z) of the circuit is given by: \[Z=\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}\] Here, \[Z=\frac{200}{0.5}=400\Omega \] Now, \[X_{L}^{2}\,={{Z}^{2}}\,-{{R}^{2}}\,={{(400)}^{2}}\,-{{(200)}^{2}}\] \[{{(2\pi /L)}^{2}}=12\times {{10}^{4}}\] \[L=\frac{2\sqrt{3}\,\times 100}{2\pi \times 50}\,=\frac{2\sqrt{3}}{\pi }\,=1.1\,H\]
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