Q.(a) Draw graphs showing the variations of inductive reactance and capacitive reactance with frequency of the applied ac source.

$X_L\propto f$ (line), $X_C\propto 1/f$ (hyperbola); X is a pure inductor, Y a pure resistor; series current $=V/\sqrt{R^2+X_L^2}=0.25/\sqrt2\approx0.18\,$A. (OR transformer: $I_p\approx5.6\,$mA.)

1. Reactance vs frequency. Inductive reactance $X_L=\omega L=2\pi f L$ — a straight line through the origin, slope $2\pi L$. Capacitive reactance $X_C=\dfrac{1}{\omega C}=\dfrac{1}{2\pi f C}$ — a rectangular hyperbola falling with $f$.
2. Series RC phasor. Take current $\vec I$ as reference. $\vec V_R$ is in phase with $I$ (along it); $\vec V_C$ lags $I$ by $90^\circ$ (drawn downward). The resultant applied voltage $\vec V=\sqrt{V_R^2+V_C^2}$ lags... in fact the current leads the applied voltage by $\phi=\tan^{-1}(V_C/V_R)$.
3. Identify devices and find current.

• X: current lags voltage by $\pi/2$ $\Rightarrow$ pure inductor. Its reactance $X_L=\dfrac{V}{I}=\dfrac{220}{0.25}=880\ \Omega.$
• Y: current in phase with voltage $\Rightarrow$ pure resistor, $R=\dfrac{220}{0.25}=880\ \Omega.$
• In series: $Z=\sqrt{R^2+X_L^2}=\sqrt{880^2+880^2}=880\sqrt2\approx1244.5\ \Omega.$ $I=\frac{V}{Z}=\frac{220}{880\sqrt2}=\frac{0.25}{\sqrt2}\approx0.177\ \text{A}\approx0.18\ \text{A}.$ …