Question:
In a common emitter transistor amplifier, AC current gain is \(65\), the load resistance is \(5400\,\Omega\) and input resistance of the transistor is \(450\,\Omega\). The voltage gain is
In a common emitter transistor amplifier, AC current gain is \(65\), the load resistance is \(5400\,\Omega\) and input resistance of the transistor is \(450\,\Omega\). The voltage gain is
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Voltage gain = current gain times (R_load / R_input).
Updated On: Oct 1, 2026
- \(460\)
- \(540\)
- \(780\)
- \(7800\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
In a common emitter amplifier, the voltage gain is \(A_V = \beta_{ac}\times\frac{R_L}{R_{in}}\), where \(\beta_{ac}\) is the ac current gain, \(R_L\) the load resistance and \(R_{in}\) the input resistance.
Step 2: Compute:
\[ A_V = 65\times\frac{5400}{450} = 65\times12 = 780 \]
Step 3: Why the other options are wrong.
460 and 540 do not match the product of 65 and 12. 7800 results from a slip of a factor of 10, such as using \(R_L = 54000\ \Omega\).
Final Answer:
The voltage gain is \(780\), option (C). \[ \boxed{780} \]
In a common emitter amplifier, the voltage gain is \(A_V = \beta_{ac}\times\frac{R_L}{R_{in}}\), where \(\beta_{ac}\) is the ac current gain, \(R_L\) the load resistance and \(R_{in}\) the input resistance.
Step 2: Compute:
\[ A_V = 65\times\frac{5400}{450} = 65\times12 = 780 \]
Step 3: Why the other options are wrong.
460 and 540 do not match the product of 65 and 12. 7800 results from a slip of a factor of 10, such as using \(R_L = 54000\ \Omega\).
Final Answer:
The voltage gain is \(780\), option (C). \[ \boxed{780} \]
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