Step 1: Recall the phase angle formula for a series RLC circuit.
In a series RLC circuit driven by an AC source, the phase angle $\phi$ between the current and the applied voltage is: \[ \tan\phi = \frac{X_L - X_C}{R} \] where $X_L$ is inductive reactance and $X_C$ is capacitive reactance.
Step 2: Calculate the inductive reactance $X_L$.
\[ X_L = \omega L = 400\,\text{rad/s} \times 500\,\text{mH} = 400 \times 0.5 = 200\,\Omega \]
Step 3: Calculate the capacitive reactance $X_C$.
\[ X_C = \frac{1}{\omega C} = \frac{1}{400 \times 20 \times 10^{-6}} = \frac{1}{8 \times 10^{-3}} = 125\,\Omega \]
Step 4: Determine the net reactance.
\[ X_L - X_C = 200 - 125 = 75\,\Omega \] Since $X_L > X_C$, the circuit is inductive and the voltage leads the current.
Step 5: Calculate $\tan\phi$.
\[ \tan\phi = \frac{75}{150} = 0.5 \]
Step 6: Find the phase angle and state the answer.
\[ \phi = \tan^{-1}(0.5) \] This means the voltage leads the current by $\tan^{-1}(0.5) \approx 26.6^\circ$. The phase angle between current and applied voltage is: \[ \boxed{\phi = \tan^{-1}(0.5)} \]