A) 5 A
B) 0.5 A
C) 0.7 A
D) 7 A
Correct Answer: B
Solution :
Wattless component of current is \[i\text{ }={{i}_{v}}\,sin\theta \] \[=\frac{{{E}_{v}}}{Z}\sin \theta \] where,\[Z=\]impedance of L-R circuit \[=\sqrt{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}\] So \[i=\frac{220}{\sqrt{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}}\sin \theta \] From impedance triangle, \[\sin \theta =\frac{L\omega }{\sqrt{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}}\] \[\therefore \] \[i=\frac{220}{\sqrt{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}}\frac{L\omega }{\sqrt{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}}\] \[=\frac{220}{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}L\omega \] \[=\frac{220\times 0.7\times 2\pi \times 50}{{{(220)}^{2}}+{{(0.7\times 2\pi \times 50)}^{2}}}\] \[=\frac{220\times 0.7\times 2\times 22/7\times 50}{{{(220)}^{2}}+{{\left( 0.7\times 2\times \frac{22}{7}\times 50 \right)}^{2}}}\] \[=\frac{220\times 220}{(220)+{{(220)}^{2}}}\] \[=0.5A\]
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