An Alternating Emf Of Peak To Peak Value Of 40 V Is Applied Across The Series Combination Of An Inductor Of Inductance 100mh Capacitor Of Capacitance 1uf And Resistance 100 Ohm . Determine The Bandwidth And Quality Factor.

Q) An Alternating Emf Of Peak To Peak Value Of 40 V Is Applied Across The Series Combination Of An Inductor Of Inductance 100mh Capacitor Of Capacitance 1uf And Resistance 100 Ohm. Determine The Bandwidth And Quality Factor.

Ans) METHOD 1:-

METHOD 2)

1. Calculate Impedance:

First, we need to determine the total impedance (Z) of the circuit at a specific frequency. Each component contributes to the impedance:

• Inductive reactance (X_L) = 2πfL, where f is the frequency.
• Capacitive reactance (X_C) = 1/(2πfC).
• Total impedance (Z) = √(R^2 + (X_L - X_C)^2).

2. Find Resonant Frequency:

At the resonant frequency (f_r), X_L and X_C cancel each other out, leaving only the resistance in the circuit. This results in minimum impedance and maximum current flow. The resonant frequency can be calculated by:

• f_r = 1/(2π√(LC)).

3. Calculate Bandwidth:

The bandwidth (B) represents the range of frequencies around the resonant frequency where the impedance remains close to its minimum value. We can estimate the bandwidth using:

• B = f_r / Q, where Q is the quality factor.

4. Compute Quality Factor:

The quality factor (Q) indicates the "sharpness" of the resonance peak. Higher Q implies a narrow peak and less energy dissipation. Q can be calculated by:

• Q = ω_rR/L, where ω_r is the angular resonant frequency (2πf_r).

5. Apply Values:

Plugging in the given values:

• L = 100 mH = 0.1 H
• C = 1 μF = 10^-6 F
• R = 100 Ω

Result:

1. Resonant Frequency:

   • f_r = 1/(2π√(0.1 * 10^-6)) ≈ 159.15 Hz

2. Bandwidth:

   • B = f_r / Q (requires Q calculation)

3. Quality Factor:

   • ω_r = 2πf_r ≈ 1000 rad/s
   • Q = ω_rR/L = 1000 * 100 / 0.1 ≈ 100000

4. Bandwidth:

   • B = 159.15 Hz / 100000 ≈ 0.00159 Hz

Therefore, the bandwidth of the circuit is approximately 0.00159 Hz, indicating a very narrow resonance peak due to the high quality factor of 100000.

Remember:

• These calculations are estimates and depend on the actual frequency dependence of the components.
• The peak-to-peak voltage of 40 V is not used in the calculations directly but provides information about the applied voltage source.

I hope this helps! Let me know if you have any further questions.