A) 50 rad/s and 5.75A
B) 1000 rad/s and 3A
C) 10 rad/s and 2A
D) 20 rad/s and 4A
Correct Answer: A
Solution :
The source frequency which drives the circuit in resonance is resonant frequency and given by \[{{\omega }_{0}}=\frac{1}{\sqrt{LC}}\] \[=\frac{1}{\sqrt{5\times 80\times {{10}^{-6}}}}=50\,rad/s\] The impedance of the circuit. \[Z=\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}\] \[Z=\sqrt{{{R}^{2}}+\left( 50\times 5-\frac{1}{50\times 80\times {{10}^{-6}}} \right)}\] \[Z=\sqrt{{{(40)}^{2}}+(250-250)}\] \[Z=\sqrt{{{(40)}^{2}}}\] \[Z=40=R\] The amplitude of current at the resonating frequency\[=V/Z=V/R=230/40=5.75\text{ }A\].You need to login to perform this action.
You will be redirected in
3 sec