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) \[=\frac{V}{R}\] \[=\frac{100}{200}=0.5A\] In a circuit containing inductive reactance \[({{X}_{L}})\]and resistance (R), impedance (Z) of the circuit is \[Z=\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}\] ...(i) Here, \[Z=\frac{200}{0.5}=400\Omega \] Now \[X_{L}^{2}={{Z}^{2}}-{{R}^{2}}\] \[={{(400)}^{2}}-{{(200)}^{2}}\] \[{{(2\pi fL)}^{2}}=12\times {{10}^{4}}\] \[L=\frac{2\sqrt{3}\times 100}{2\pi \times 50}\] \[=\frac{2\sqrt{3}}{\pi }=1.1H\]
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