Physics - EMI/AC Question with Solution | TestHub

In the first case, capacitive reactance, $X_C=\frac{1}{\omega C}$ Impedance of the circuit, $Z=\sqrt{R^2+X_C^2}$ Current in the circuit, $I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+X_C^2}}\dots$ (i) In the second case, ${\omega }^{\prime }=\frac{\omega }{3}$ $\therefore$ Capacitive reactance, $X_C^{\prime }=\frac{1}{{\omega }^{\prime }C}=\frac{1}{\frac{\omega }{3}C}=\frac{3}{\omega C}=3X_C$ Impedance of the circuit, $Z^{\prime }=\sqrt{R^2+X_C^{\prime 2}}=\sqrt{R^2+9X_C^2}$ Current in the circuit, $\frac{I}{2}=\frac{V}{Z^{\prime }}=\frac{V}{\sqrt{R^2+9X_C^2}}$ Divide (ii) by (i), we get $\begin{array}{rl}& \frac{1}{2}=\frac{\sqrt{R^2+X_C^2}}{\sqrt{R^2+9X_C^2}}\Rightarrow \frac{1}{4}=\frac{R^2+X_C^2}{R^2+9X_C^2}\\ & 5X_C^2=3R^2\text{ or }\frac{X_C}{R}=\sqrt{\frac{3}{5}}=\sqrt{0.6}\end{array}$