In a transistor connected in a common emitter mode `R_C = 4kOmega, R_1 = 1kOmega, I_C = 1mA` and `I_B = 20 muA`. Find the voltage gain.

Options

A. 100
B. 200
C. 300
D. 400

Explanation

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

### Step 1: Calculate the Current Gain (β)
The current gain (β) of a transistor is given by the formula:
\[ \beta = \frac{I_C}{I_B} \]
Where:
- \( I_C = 1 \, \text{mA} = 1000 \, \mu\text{A} \)
- \( I_B = 20 \, \mu\text{A} \)

Substituting the values:
\[ \beta = \frac{1000 \, \mu\text{A}}{20 \, \mu\text{A}} = 50 \]

### Step 2: Calculate the Resistance Gain (R_gain)
The resistance gain is calculated using the formula:
\[ R_{\text{gain}} = \frac{R_C}{R_1} \]
Where:
- \( R_C = 4 \, k\Omega \)
- \( R_1 = 1 \, k\Omega \)

Substituting the values:
\[ R_{\text{gain}} = \frac{4 \, k\Omega}{1 \, k\Omega} = 4 \]

### Step 3: Calculate the Voltage Gain (A_V)
The voltage gain (A_V) is the product of the current gain and the resistance gain:
\[ A_V = \beta \times R_{\text{gain}} \]

Substituting the values:
\[ A_V = 50 \times 4 = 200 \]

### Final Answer
The voltage gain of the transistor in common emitter mode is:
\[ \text{Voltage Gain} = 200 \]