A transistor is used in Common-emitter mode in an amplifier circuits. When a signal of 20 mV is added to the base-emitter voltage, the base current changes by `40 muA` and the collector current changes by 2mA. The load resistance is `5 kOmega` then the voltage gain is

Options

A. 500
B. 200
C. 400
D. 1000

Answer

To find the voltage gain of the transistor amplifier in common-emitter mode, we can follow these steps:

### Step 1: Identify the given values
- Change in base current, \( \Delta I_B = 40 \, \mu A = 40 \times 10^{-6} \, A \)
- Change in collector current, \( \Delta I_C = 2 \, mA = 2 \times 10^{-3} \, A \)
- Load resistance, \( R_L = 5 \, k\Omega = 5 \times 10^{3} \, \Omega \)
- Input signal voltage, \( V_{in} = 20 \, mV = 20 \times 10^{-3} \, V \)

### Step 2: Calculate the output voltage
The output voltage \( V_{out} \) can be calculated using the formula:
\[ V_{out} = R_L \times \Delta I_C \]
Substituting the values:
\[ V_{out} = 5 \times 10^{3} \, \Omega \times 2 \times 10^{-3} \, A = 10 \, V \]

### Step 3: Calculate the voltage gain
The voltage gain \( A_V \) is defined as the ratio of the output voltage to the input voltage:
\[ A_V = \frac{V_{out}}{V_{in}} \]
Substituting the values:
\[ A_V = \frac{10 \, V}{20 \times 10^{-3} \, V} = \frac{10}{0.02} = 500 \]

### Conclusion
The voltage gain of the transistor amplifier in common-emitter mode is \( 500 \).