A) 3.33 A
B) 33.3 A
C) 10 A
D) 5 A
Correct Answer: A
Solution :
The inductive reactance\[{{X}_{L}}=\omega L=2\pi f\,L\] Given, \[L=0.2H,\,\,f=50\,Hz\] \[\therefore \] \[{{X}_{L}}=2\times \pi \times 50\times 0.2\] and \[R=20\,\,\Omega \] Hence, impedance \[Z=\sqrt{{{X}_{L}}^{2}+{{R}^{2}}}\] \[\Rightarrow \] \[Z=\sqrt{{{(2\pi \times 50\times 0.2)}^{2}}+{{(20)}^{2}}}\] and current \[I=\frac{V}{Z}=\frac{220}{\sqrt{3943.84+440}}\] \[\Rightarrow \] \[I=\frac{220}{65.9}=3.33\,A\]You need to login to perform this action.
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