A) \[R=1k\Omega ,\,C=10\,\mu F\]
B) \[R=1k\Omega ,\,C=1\,\mu F\]
C) \[R=1k\Omega ,\,L=1\,0\,mH\]
D) \[R=10k\Omega ,\,L=1\,0\,mH\]
Correct Answer: A
Solution :
(a) \[R=1k\Omega ,\,C=10\,\mu F\] Figure given in the question shows that current l leads the voltage V by a phase angle \[\pi /4\]. Therefore, the circuit can be RC circuit alone. \[\tan \,\phi =\frac{{{X}_{C}}}{R}=\frac{1}{\omega CR}\left( \because \,{{X}_{C}}=\frac{1}{\omega C} \right)\] \[\tan \frac{\pi }{4}=\frac{1}{\omega CR};1=\frac{1}{\omega CR}\] (1) From \[V={{V}_{0}}\sin \,100\,t\]. we get, \[\omega =100\,rad\,{{s}^{-1}}\] \[\therefore CR=\frac{1}{\omega }=\frac{1}{100}\] (Using eq. (1)) When \[R=1\,k\Omega ={{10}^{3}}\Omega ,\,C=\frac{1}{{{10}^{5}}}={{10}^{-5}}F=10\,\mu F\]
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