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 $Z = \frac{2R}{\sqrt{5}}\quad\therefore I_{0} = \frac{V_{0}}{Z} = \frac{\sqrt{5}V_{0}}{2R}$