An inductor of $\left(\frac{100}{\pi}\right) \mathrm{mH}$, capacitor of capacitance $\left(\frac{10^{-3}}{2…

The tangent of the phase angle between voltage and current in a series RLC circuit is determined by the inductive and capacitive reactances relative to the resistance.

Given: $R = 10 \Omega$, $L = \frac{100}{\pi} \times 10^{-3} \,\mathrm{H}$, $C = \frac{10^{-3}}{2\pi} \,\mathrm{F}$, and $f = 50 \,\mathrm{Hz}$.

The angular frequency is $\omega = 2\pi f = 100\pi \mathrm{rad/s}$.

Inductive reactance is $X_L = \omega L = (100\pi) \cdot \left(\frac{100}{\pi} \times 10^{-3}\right) = 10 \,\Omega$.

Capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100\pi \cdot \frac{10^{-3}}{2\pi}} = \frac{1}{0.05} = 20 \,\Omega$.

The phase angle satisfies $\tan\phi = \frac{X_L - X_C}{R} = \frac{10 - 20}{10} = -1$. Taking the magnitude gives $|\tan\phi| = 1$.

The tangent of the phase angle is $1$.