The input resistance of a silicon transistor is `100 Omega`. Base current is changed by `40 muA` which results in a change in collector current by `2 mA`. This transistor is used as a common-emitter amplifier with a load resistance of `4k Omega`. The voltage gain of the amplifier is

Options

A. 3000
B. 4000
C. 1000
D. 2000

Explanation

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

### Step 1: Calculate Beta (β)
Beta (β) is defined as the ratio of the change in collector current (I_C) to the change in base current (I_B).

Given:
- Change in base current, I_B = 40 µA = 40 × 10^(-6) A
- Change in collector current, I_C = 2 mA = 2 × 10^(-3) A

Using the formula:
\[ \beta = \frac{I_C}{I_B} \]

Substituting the values:
\[ \beta = \frac{2 \times 10^{-3}}{40 \times 10^{-6}} = \frac{2 \times 10^{-3}}{4 \times 10^{-5}} = 50 \]

### Step 2: Calculate the Voltage Gain (A_v)
The voltage gain (A_v) of a common-emitter amplifier can be calculated using the formula:
\[ A_v = \beta \times \frac{R_L}{R_{in}} \]
where:
- \(R_L\) = Load resistance
- \(R_{in}\) = Input resistance

Given:
- Load resistance, \(R_L = 4 k\Omega = 4000 \Omega\)
- Input resistance, \(R_{in} = 100 \Omega\)

Substituting the values:
\[ A_v = 50 \times \frac{4000}{100} \]
\[ A_v = 50 \times 40 = 2000 \]

### Final Answer:
The voltage gain of the amplifier is **2000**.

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