Question

A $n-p-n$ transistor is connected to common emitter configuration in a given amplifier. A load resistance of $800$ is connected in the collector circuit and the voltage drop across it is $0.8\, V$. If the current amplification factor is $0.96$ and the input resistance of the circuit is $192\,\Omega$, the voltage gain and the power gain of the amplifier will respectively be

• A. 3.69, 3.84
• B. 4 , 4
• C. 4, 3.69
• D. 4, 3.84

Answer 1

The Correct Option is D

To solve this problem, we need to calculate the voltage gain and power gain of the amplifier, given the parameters.

Step 1: Understanding the Problem

We have an n-p-n transistor amplifier in a common emitter configuration. The details are:

• Load resistance \( R_L = 800 \, \Omega \)
• Voltage drop across the load resistance \( V_{RC} = 0.8 \, V \)
• Current amplification factor \( \beta = 0.96 \)
• Input resistance \( R_i = 192 \, \Omega \)

Step 2: Calculating Voltage Gain

The voltage gain \( A_V \) in a common emitter configuration can be calculated using the formula:

A_V = \beta \times \left(\frac{R_L}{R_i}\right)

Substituting the given values:

A_V = 0.96 \times \left(\frac{800}{192}\right)

Calculating further:

A_V = 0.96 \times 4.1667 = 4

Step 3: Calculating Power Gain

The power gain \( A_P \) can be calculated using:

A_P = A_V \times \beta

Substitute the calculated voltage gain and the given current amplification factor:

A_P = 4 \times 0.96

Calculating further:

A_P = 3.84

Conclusion

Therefore, the voltage gain is 4 and the power gain is 3.84. Hence, the correct answer is the option 4, 3.84.