1. Given data

• Resistance: $R = 5\,\Omega$
• Inductance: $L = 25\text{ mH} = 25 \times 10^{-3}\text{ H}$
• Capacitance: $C = 1000\,\mu\text{F} = 1000 \times 10^{-6}\text{ F} = 10^{-3}\text{ F}$
• Angular frequency: $\omega = 320\,\text{s}^{-1}$

We need the total impedance and phase difference for a series LCR circuit.

2. Inductive reactance

$X_L = \omega L = 320 \times 25 \times 10^{-3} = 8\,\Omega$

3. Capacitive reactance

$X_C = \frac{1}{\omega C} = \frac{1}{320 \times 10^{-3}} = \frac{1}{0.32} = 3.125\,\Omega$

4. Net reactance

For a series LCR circuit,

$X = X_L - X_C = 8 - 3.125 = 4.875\,\Omega$

Since $X_L > X_C$, the circuit is overall inductive.

5. Total impedance

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

$Z = \sqrt{5^2 + 4.875^2} = \sqrt{25 + 23.765625} = \sqrt{48.765625} \approx 6.98\,\Omega$

So,

$Z \approx 7\,\Omega$

6. Phase difference

For a series LCR circuit,

$\tan \phi = \frac{X_L - X_C}{R} = \frac{4.875}{5} = 0.975$

Thus,

$\phi = \tan^{-1}(0.975) \approx 44.3^\circ \approx 45^\circ$

So the voltage leads the current by about $45^\circ$.

7. Check options

• A: $10\,\Omega$ and $\tan^{-1}(5/3)$ → incorrect impedance
• B: $7\,\Omega$ and $45^\circ$ → matches
• C: $10\,\Omega$ and $\tan^{-1}(8/3)$ → incorrect
• D: $7\,\Omega$ and $\tan^{-1}(5/3)$ → incorrect phase angle

8. Final answer

The total impedance and phase difference are:

$\boxed{7\,\Omega \text{ and } 45^\circ}$

So the correct option is B.