Step 1: Understanding the Concept:
In a series RL circuit, the current and voltage are out of phase because the inductor causes the voltage to lead the current.
We need to find this phase angle by comparing the inductive reactance to the resistance.
Step 2: Key Formula or Approach:
Inductive reactance $X_L = \omega L = 2\pi f L$.
Phase difference $\phi$ is given by $\tan \phi = \frac{X_L}{R}$.
Step 3: Detailed Explanation:
Given $R = 450 \Omega$, $L = 1.5 \text{ H}$, and $f = \frac{150}{\pi} \text{ Hz}$.
First, find angular frequency $\omega$:
\[ \omega = 2\pi f = 2\pi \left(\frac{150}{\pi}\right) = 300 \text{ rad/s} \]
Next, compute the inductive reactance $X_L$:
\[ X_L = \omega L = 300 \times 1.5 = 450 \Omega \]
Calculate the phase angle tangent:
\[ \tan \phi = \frac{X_L}{R} = \frac{450}{450} = 1 \]
Taking the inverse tangent gives the phase difference:
\[ \phi = \tan^{-1}(1) \]
Step 4: Final Answer:
The phase difference is $\tan^{-1}(1)$.