Question

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

• A. $\frac{\sqrt{5}{V}_0}{2R}$
• B. $\frac{V_0}{2\sqrt{2}R}$
• C. $\frac{V_0}{2R}$
• D. $\frac{V_0}{2\sqrt{3}R}$

Answer 1

Answer: (A)

$\frac{1}{Z}=\sqrt{\frac{1}{R^2}+{\left(\frac{1}{X_C}-\frac{1}{X_L}\right)}^2}$
Substituting the given values, we get
$\frac{1}{Z}=\sqrt{\frac{1}{R^2}+{\left(\frac{1}{2R}-\frac{1}{R}\right)}^2}$
$\frac{1}{Z}=\sqrt{\frac{1}{R^2}+\frac{1}{4R^2}}$
$\frac{1}{Z}=\frac{\sqrt{5}}{2R}$ or $Z=\frac{2R}{\sqrt{5}}\therefore I_0=\frac{V_0}{Z}=\frac{\sqrt{5}{V}_0}{2R}$