Question:
In the circuit shown, \(R_1 = 100 \, \text{k}\Omega\) and \(R_2 = 1 \, \text{k}\Omega\). If the base-to-emitter voltage of the npn BJT is 0.7 V and the collector-to-emitter voltage is 5.2 V, the \(\beta\) (current gain) of the BJT is _________ (round off to two decimal places).
In the circuit shown, \(R_1 = 100 \, \text{k}\Omega\) and \(R_2 = 1 \, \text{k}\Omega\). If the base-to-emitter voltage of the npn BJT is 0.7 V and the collector-to-emitter voltage is 5.2 V, the \(\beta\) (current gain) of the BJT is _________ (round off to two decimal places).
Updated On: Nov 25, 2025
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Correct Answer: 148
Solution and Explanation
The current gain \( \beta \) is given by:
\[
\beta = \frac{I_C}{I_B}
\]
First, calculate the base current \(I_B\) using KVL for the base loop:
\[
I_B = \frac{12V - 0.7V}{100 \times 10^3 \, \Omega} = 113 \, \mu A
\]
Now, calculate the collector current \(I_C\) using KVL for the collector loop:
\[
I_C = \frac{12V - 5.2V}{1 \times 10^3 \, \Omega} = 6.8 \, mA
\]
Thus, the current gain is:
\[
\beta = \frac{6.8 \, mA}{113 \, \mu A} = 60.18
\]
Therefore, the value of \(\beta\) is \( \boxed{60.18} \).
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