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).
The magnitude of the current gain \(\frac{I_{\text{load}}}{I_{\text{in}}}\) in the circuit below is ______ (rounded off to two decimal places).
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When dependent sources appear, find controlling current first, then treat the dependent source as a fixed injection source for the load network.
Updated On: Dec 4, 2025
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Correct Answer: 75.49
Solution and Explanation
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
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