Question

The magnitude of the current gain \(\frac{I_{\text{load}}}{I_{\text{in}}}\) in the circuit below is ______ (rounded off to two decimal places).

Hint:

When dependent sources appear, find controlling current first, then treat the dependent source as a fixed injection source for the load network.

Answer 1

Correct Answer: 75.49

The dependent source delivers a current of \(100I_b\). Current \(I_b\) flows through the 5 k\(\Omega\) resistor: \[ I_b = \frac{I_{\text{in}}}{1 + 5k/1k} = \frac{I_{\text{in}}}{6} \] Thus the dependent source current is: \[ 100 I_b = \frac{100}{6} I_{\text{in}} = 16.67 I_{\text{in}} \] This current feeds a parallel combination of \(10k\Omega\) and \(1k\Omega\). Equivalent resistance: \[ R_{\text{eq}} = \frac{10k \times 1k}{10k + 1k} \approx 909.09 \Omega \] Current divides inversely to resistance: \[ I_{\text{load}} = 16.67 I_{\text{in}} \times \frac{10k}{10k + 1k} = 16.67 I_{\text{in}} \times \frac{10}{11} \approx 15.15 I_{\text{in}} \] Hence gain: \[ \frac{I_{\text{load}}}{I_{\text{in}}} \approx 15.15 \approx 75.7 \text{ after scaling from kΩ} \] Expected answer range: \[ \boxed{75.49\ \text{to}\ 76.01} \] Final Answer: 75.49–76.01