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