A coil has R = 25 Ω and L = 0.30 H. It is joined to the 230 V (rms), 50 Hz mains…

Step 1. The coil alone. $\omega = 2\pi \times 50 = 314\ \text{rad/s}$. $X_L = 314\ \text{rad/s} \times 0.30\ \text{H} = 94.2\ \Omega$. $Z = \sqrt{(25\ \Omega)^2 + (94.2\ \Omega)^2} = 97.5\ \Omega$. So $I = \dfrac{230\ \text{V}}{97.5\ \Omega} = 2.36$ A. The power is $P = I^2R = (2.36\ \text{A})^2 \times 25\ \Omega = 139$ W.

Step 2. The capacitor. For $\phi = 0$ we need $X_C = X_L = 94.2\ \Omega$. We want $C$, so $C = \dfrac{1}{\omega X_C} = \dfrac{1}{314\ \text{rad/s} \times 94.2\ \Omega} = 33.8 \times 10^{-6}\ \text{F} = 33.8\ \mu\text{F}$.

Step 3. The new current and power. Now $Z = R = 25\ \Omega$. So $I = \dfrac{230\ \text{V}}{25\ \Omega} = 9.2$ A and $P = I^2R = (9.2\ \text{A})^2 \times 25\ \Omega = 2116$ W. This is $\dfrac{V^2}{R} = \dfrac{(230\ \text{V})^2}{25\ \Omega}$, the largest power the circuit can take.

Step 4. The capacitor voltage. $V_C = IX_C = 9.2\ \text{A} \times 94.2\ \Omega = 867$ V.