A

Step 1: Calculate the inductive reactance (X_L) of the relay coil.
The formula for inductive reactance is given by:
$$ X_L = 2 \pi f L $$
where:
f = frequency = 800 cps (or 800 Hz),
L = inductance = 8 mH = 8 \times 10^{-3} H.

Substituting the values,
$$ X_L = 2 \pi \times 800 \times 8 \times 10^{-3} $$
$$ = 2 \pi \times 6.4 \approx 40.25 \Omega $$.

Step 2: Calculate the total impedance (Z) of the coil.
The total impedance for a circuit with resistance and inductive reactance is
$$ Z = \sqrt{R^2 + X_L^2} $$
where R = resistance = 30 ohms.

Therefore,
$$ Z = \sqrt{30^2 + 40.25^2} $$
$$ = \sqrt{900 + 1620.0625} $$
$$ = \sqrt{2520.0625} \approx 50.20 \Omega $$.

Step 3: Calculate the current (I) through the coil.
Using Ohm's law,
$$ I = \frac{V}{Z} $$
where V = applied voltage = 5 volts.

Substituting in the values,
$$ I = \frac{5}{50.20} \approx 0.0995 A \text{ or } 99.5 mA $$.

Step 4: Calculate the phase angle (φ) of the current relative to the voltage.
The phase angle can be calculated using:
$$ \tan(\phi) = \frac{X_L}{R} $$
Substituting values,
$$ \tan(\phi) = \frac{40.25}{30} \approx 1.3417 $$
Now using the arctangent function,
$$ \phi = \tan^{-1}(1.3417) \approx 53.06^\circ $$.

Step 5: Reducing the phase angle to zero.
To achieve a phase angle of zero, one could add a capacitor in parallel with the coil. The capacitor should have a reactance that matches the inductive reactance:
$$ X_C = -X_L $$.
Therefore, to maintain the same current, one must ensure that the capacitor reactance counterbalances the inductor's reactance, effectively canceling out the phase shift caused by the inductance.