To determine the current in the circuit and the phase angle between the current and voltage, follow these steps:

Components and their Reactance:

Inductive reactance, $ X_L = 45 \, \Omega $

Capacitive reactance, $ X_C = 25 \, \Omega $

Resistance, $ R = 20 \, \Omega $

Calculate Current (I):

The current through the series circuit can be calculated using the formula:

$ I = \frac{V}{\sqrt{(X_L - X_C)^2 + R^2}} $

Substituting the given values:

$ I = \frac{220}{\sqrt{(45 - 25)^2 + 20^2}} = \frac{220}{\sqrt{20^2 + 20^2}} $

Simplifying further:

$ I = \frac{220}{\sqrt{400 + 400}} = \frac{220}{\sqrt{800}} = \frac{220}{2\sqrt{200}} = \frac{220}{2\sqrt{100 \times 2}} = \frac{220}{20\sqrt{2}} = \frac{11}{\sqrt{2}} $

Therefore, the current is approximately:

$ I = 7.779 \, \text{A} \, (\approx 7.8 \, \text{A}) $

Calculate Phase Angle ($\phi$):

The phase angle can be calculated using the tangent of the angle:

$ \tan \phi = \frac{X_L - X_C}{R} = \frac{45 - 25}{20} = \frac{20}{20} = 1 $

Hence, the phase angle $\phi$ is:

$ \phi = 45^{\circ} $

Thus, the current in the circuit is approximately 7.8 A, and the phase angle is $45^{\circ}$.