A) 1 amp
B) 1.5 amp
C) 2 amp
D) 2.5 amp
Correct Answer: C
Solution :
According to the question. \[\tan {{60}^{o}}=\frac{\omega L}{R}\]and \[\tan {{60}^{o}}=\frac{1/\omega c}{R}\] \[\therefore \]\[\omega L=\frac{1}{\omega c},\](case of resonance) Now, \[z=\sqrt{{{R}^{2}}+\left( \omega L-\frac{1}{\omega c} \right)}=R\] \[\therefore \]\[{{I}_{rms}}=\frac{{{E}_{rms}}}{z}=\frac{{{E}_{rms}}}{R}=\frac{200v}{100\Omega }=2\,Amp.\] Hence, the correction option is [c].You need to login to perform this action.
You will be redirected in
3 sec