A resistor, an inductor and a capacitor are joined in series to an alternating source, as in the…

(a) Impedance. $X_L = \omega L = 500\ \text{rad/s} \times 0.20\ \text{H} = 100\ \Omega$. $X_C = \dfrac{1}{\omega C}$, with $C = 25 \times 10^{-6}\ \text{F}$, which is $80\ \Omega$. The inductive reactance is larger, so $X = X_L - X_C = 20\ \Omega$. $Z = \sqrt{R^2 + X^2} = \sqrt{(15\ \Omega)^2 + (20\ \Omega)^2} = 25.0\ \Omega$.

(b) Current. $I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{200\ \text{V}}{25.0\ \Omega} = 8.0\ \text{A}$.

(c) Power. Only the resistor uses power: $P = I_{rms}^2 R = (8.0\ \text{A})^2 \times 15\ \Omega = 960\ \text{W}$.

(d) Power factor. $\cos\phi = \dfrac{R}{Z} = \dfrac{15\ \Omega}{25.0\ \Omega} = 0.60$. The phase angle is $53^\circ$, and the voltage leads the current because the circuit is inductive.

(e) Second capacitor. Resonance at $\omega = 500\ \text{rad/s}$ needs $C_{total} = \dfrac{1}{\omega^2 L} = \dfrac{1}{(500\ \text{rad/s})^2 \times 0.20\ \text{H}} = 20\ \mu\text{F}$. This is smaller than $25\ \mu\text{F}$, so the new capacitor goes in series: $\dfrac{1}{20\ \mu\text{F}} = \dfrac{1}{25\ \mu\text{F}} + \dfrac{1}{C_2}$, which gives $C_2 = 100\ \mu\text{F}$.