Topic: Network Theory
The voltage V in the circuit given below is :
- A. 25 Mrad/s, 250 Krad/s, 100
- B. 5 Mrad/s, 250 Krad/s, 20
- C. 5 Mrad/s, 25 Krad/s, 200
- D. None of the above
Answer: B. 5 Mrad/s, 250 Krad/s, 20
Explanation: First, convert the circuit to a standard parallel RLC form by combining the resistors. The equivalent parallel resistance is $R_p = 1.25 M\Omega || 5 M\Omega = \frac{1.25 \times 5}{1.25 + 5} = 1 M\Omega$.
Given $L = 10 mH$ and $C = 4 pF$.
Resonant frequency: $\omega_o = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10\times 10^{-3} \times 4\times 10^{-12}}} = \frac{1}{\sqrt{40\times 10^{-15}}} = 5 \times 10^6$ rad/s = 5 Mrad/s.
Quality factor: $Q = \omega_o R_p C = (5\times 10^6) \times (1\times 10^6) \times (4\times 10^{-12}) = 20$.
Bandwidth: $\beta = \frac{\omega_o}{Q} = \frac{5\times 10^6}{20} = 0.25\times 10^6$ rad/s = 250 Krad/s.