Question & Answer

1. Calculate the resonant frequency $\omega_0$ of the circuit considering the non-ideal resistances $R_L$ and $R_C$.

The resonant frequency $\omega_0$ of an RLC circuit is given by:

$$\omega_0 = \frac{1}{\sqrt{LC}}$$

However, in this case, we also need to consider the non-ideal resistances $R_L$ and $R_C$. Since these resistances do not affect the resonant frequency directly, we calculate $\omega_0$ using the formula for an ideal RLC circuit.

Given:

• $L = 10\, \text{mH} = 10 \times 10^{-3}\, \text{H}$
• $C = 1\, \mu\text{F} = 1 \times 10^{-6}\, \text{F}$

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(10 \times 10^{-3})(1 \times 10^{-6})}}$$

Let's calculate this.

$$\omega_0 = \frac{1}{\sqrt{10 \times 10^{-9}}} = \frac{1}{\sqrt{10^{-8}}} = \frac{1}{10^{-4}} = 10^4 = 10,000 \, \text{rad/s}$$

So, the resonant frequency $\omega_0$ is 10,000 rad/s.

2. Determine the quality factor $Q$ of the circuit at resonance.

The quality factor $Q$ for a series RLC circuit with non-ideal resistances can be calculated as:

$$Q = \frac{\omega_0 L}{R_{\text{total}}}$$

where $R_{\text{total}} = R + R_L + R_C$.

Given:

• $R = 50\, \Omega$
• $R_L = 5\, \Omega$
• $R_C = 1\, \Omega$
• $L = 10 \times 10^{-3}\, \text{H}$
• $\omega_0 = 10,000\, \text{rad/s}$

$$R_{\text{total}} = 50 + 5 + 1 = 56\, \Omega$$

Now, calculate $Q$:

$$Q = \frac{\omega_0 L}{R_{\text{total}}} = \frac{(10,000)(10 \times 10^{-3})}{56}$$

So, the quality factor $Q$ is approximately 1.79.

3. Calculate the total impedance $Z(\omega)$ of the circuit at the resonant frequency $\omega_0$.

At resonance, the inductive reactance $X_L$ and capacitive reactance $X_C$ cancel each other out, so the impedance $Z$ is purely resistive and is the sum of all resistances:

$$Z(\omega_0) = R + R_L + R_C = 56\, \Omega$$

So, the total impedance $Z(\omega_0)$ at the resonant frequency is 56 Ω.

4. Compute the current amplitude $I_0$ at resonance.

The current amplitude $I_0$ at resonance can be calculated using Ohm's law:

$$I_0 = \frac{V_0}{Z(\omega_0)}$$

Given:

• $V_0 = 10\, \text{V}$
• $Z(\omega_0) = 56\, \Omega$

$$I_0 = \frac{10}{56} \approx 0.179 \, \text{A}$$

So, the current amplitude $I_0$ at resonance is approximately 0.179 A.

5. Discuss the impact of the non-ideal resistances on the circuit's performance compared to an ideal RLC circuit.

In an ideal RLC circuit (where $R_L = 0$ and $R_C = 0$), the only resistance would be $R = 50\, \Omega$. This would result in a higher quality factor $Q$, which means a narrower bandwidth around the resonant frequency, indicating a more selective and sharper resonance peak.

However, with non-ideal resistances $R_L$ and $R_C$, the total resistance increases, reducing the quality factor $Q$ to about 1.79, and increasing the impedance $Z(\omega_0)$ at resonance. This causes the resonance peak to be broader and less sharp, resulting in less selectivity. Additionally, the higher resistance leads to a lower current amplitude $I_0$ for a given input voltage $V_0$.

Overall, the non-ideal resistances degrade the performance of the circuit by reducing the sharpness and selectivity of the resonance and decreasing the current amplitude at resonance.