A) 5 A
B) 0.5 A
C) 0.7 A
D) 7 A
Correct Answer: B
Solution :
Watt less component of current \[={{I}_{v}}\sin \theta \] \[=\frac{{{E}_{v}}}{Z}\sin \theta \] \[=\frac{220}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}\times \frac{\omega L}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}\] \[=\frac{220\times \omega L}{({{R}^{2}}+{{\omega }^{2}}{{L}^{2}})}\] \[=\frac{220\times (2\pi \times 50\times 0.7)}{{{(220)}^{2}}+{{(2\pi \times 50\times 0.7)}^{2}}}\] \[=\frac{220\times 220}{{{(220)}^{2}}+{{(220)}^{2}}}\] \[=\frac{1}{2}\] \[=0.5\text{ }A\]
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