A) 0.69 H
B) 0.28 H
C) 0.38 H
D) 0.56 H
Correct Answer: B
Solution :
[b] For power to be consumed at the rate of \[\frac{1100}{5}=220W,\]we have \[P={{E}_{v}}{{I}_{v}}\cos \theta \] \[220=\frac{220\times 220}{\sqrt{{{R}^{2}}+{{L}^{2}}}{{\omega }^{2}}}\times \frac{R}{\sqrt{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}}\] Where \[R=\frac{{{V}^{2}}}{P}=\frac{{{220}^{2}}}{1100}=44\Omega \] \[220=\frac{{{(220)}^{2}}\times 44}{{{44}^{2}}+{{(L\omega )}^{2}}};{{44}^{2}}+{{(L\omega )}^{2}}=220\] \[{{(L\omega )}^{2}}=\sqrt{220\times 44-{{44}^{2}}}\]\[=\sqrt{44(220-44)}=\sqrt{44\times 176}=88\Omega \] \[L=\frac{88}{2\pi \times f}=\frac{88}{2\pi \times 50}=\frac{88}{2\times 22}\times \frac{7}{50}=0.28H\]
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