A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[0.5\]
D) \[0.6\]
Correct Answer: A
Solution :
: Here, \[R=500\Omega \], \[L=0.5H\] Compare \[V=100\sqrt{2}\sin (1000t)\]with \[V={{V}_{0}}\sin \omega t\] we get \[\omega =1000\] The inductive reactance is \[{{X}_{L}}=\omega L=(1000)(0.5)=500\Omega \] Impedance of the RL circuit is \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}=\sqrt{{{(500\Omega )}^{2}}+{{(500\Omega )}^{2}}}500\sqrt{2}\Omega \]Power factor, \[\cos \phi =\frac{R}{Z}=\frac{500\Omega }{500\sqrt{2}\Omega }=\frac{1}{\sqrt{2}}\]
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