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 In the circuit shown in figure \mathrm{X_L=\frac{X_C}{2}=R} the peak value current \mathrm{ I_0}  is: 
                                                                          
                    

Option: 1

\frac{\sqrt{5} \mathrm{~V}_0}{2 \mathrm{R}}


Option: 2

\frac{V_0}{2 \sqrt{2} R}


Option: 3

\frac{\mathrm{V}_0}{2 \mathrm{R}}


Option: 4

\frac{V_0}{2 \sqrt{3} R}


Answers (1)

best_answer


\mathrm{\frac{1}{Z}=\sqrt{\frac{1}{R^2}+\left(\frac{1}{X_C}-\frac{1}{X_L}\right)} }
Substituting the given values, we get
\mathrm{\begin{aligned} & \frac{1}{Z}=\sqrt{\frac{1}{R^2}+\left(\frac{1}{2 R}-\frac{1}{R}\right)^2} \\ & \frac{1}{Z}=\sqrt{\frac{1}{R^2}+\frac{1}{4 R^2}} \\ & \frac{1}{Z}=\frac{\sqrt{5}}{2 R} \quad \text { or } \quad Z=\frac{2 R}{\sqrt{5}} \quad \therefore \quad I_0=\frac{V_0}{Z}=\frac{\sqrt{5} V_0}{2 R} \end{aligned} }

Posted by

Ajit Kumar Dubey

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