This is an RLC circuit. To solve the problem, let's analyze and compute each parameter step by step:

1. Total Impedance of the Circuit at Resonance:

At resonance, the inductive reactance ($X_L$) and capacitive reactance ($X_C$) cancel each other. The total impedance of the circuit will then be equal to the resistance $R$ in the circuit.

Given:

• $R = 25 \, \Omega$

At resonance: $Z_{\text{total}} = R = 25 \, \Omega$

2. Resonant Frequency ($f_r$):

The resonant frequency is determined using the formula: $f_r = \frac{1}{2\pi \sqrt{L C}}$ Where:

• $L = 0.5 \, \text{H}$
• $C = 5 \, \mu\text{F} = 5 \times 10^{-6} \, \text{F}$

Substituting: $f_r = \frac{1}{2\pi \sqrt{(0.5)(5 \times 10^{-6})}}$ $f_r = \frac{1}{2\pi \sqrt{2.5 \times 10^{-6}}}$ $f_r = \frac{1}{2\pi \times 0.0015811} \approx 100.8 \, \text{Hz}$

3. Bandwidth ($BW$):

The bandwidth is calculated as: $BW = \frac{R}{2\pi L}$ Substituting: $BW = \frac{25}{2\pi (0.5)} = \frac{25}{3.1416} \approx 7.96 \, \text{Hz}$

4. Quality Factor ($Q$):

The quality factor is determined as: $Q = \frac{f_r}{BW}$ Substituting the values: $Q = \frac{100.8}{7.96} \approx 12.66$

Final Answers:

1. Total Impedance at Resonance: $25 \, \Omega$
2. Resonant Frequency: $100.8 \, \text{Hz}$
3. Bandwidth: $7.96 \, \text{Hz}$
4. Quality Factor: $12.66$

Would you like a detailed explanation of any of these steps or related visualizations?

Here are 5 related questions you might consider:

1. How does the quality factor affect the sharpness of resonance?
2. What would happen to the resonant frequency if the inductance was doubled?
3. How can we modify the bandwidth of this circuit?
4. What is the role of the resistor in determining the circuit’s behavior?
5. How does a change in the capacitance value influence the circuit's resonance?

Tip: Always double-check the units of $L$, $C$, and $R$ to ensure accuracy in your calculations!