In a transistor, the change in base current from `100 muA` to`125 muA `causes a change in collector current from 5 mA to 7.5 mA, keeping collector-to-emitter voltage constant at 10V. What is the current gain of the transistor?

Options

A. 200
B. 100
C. 50
D. 25

Explanation

To find the current gain (β) of the transistor, we can follow these steps:

### Step 1: Identify the change in collector current (ΔIC)
The change in collector current (ΔIC) is calculated as:
\[ \Delta I_C = I_{C(final)} - I_{C(initial)} \]
Given:
- \(I_{C(initial)} = 5 \, \text{mA}\)
- \(I_{C(final)} = 7.5 \, \text{mA}\)

Calculating ΔIC:
\[ \Delta I_C = 7.5 \, \text{mA} - 5 \, \text{mA} = 2.5 \, \text{mA} \]

### Step 2: Identify the change in base current (ΔIB)
The change in base current (ΔIB) is calculated as:
\[ \Delta I_B = I_{B(final)} - I_{B(initial)} \]
Given:
- \(I_{B(initial)} = 100 \, \mu A = 100 \times 10^{-6} \, A\)
- \(I_{B(final)} = 125 \, \mu A = 125 \times 10^{-6} \, A\)

Calculating ΔIB:
\[ \Delta I_B = 125 \times 10^{-6} \, A - 100 \times 10^{-6} \, A = 25 \times 10^{-6} \, A \]

### Step 3: Calculate the current gain (β)
The current gain (β) is defined as the ratio of the change in collector current to the change in base current:
\[ \beta = \frac{\Delta I_C}{\Delta I_B} \]
Substituting the values we calculated:
\[ \beta = \frac{2.5 \times 10^{-3} \, A}{25 \times 10^{-6} \, A} \]

### Step 4: Simplify the expression
To simplify:
\[ \beta = \frac{2.5 \times 10^{-3}}{25 \times 10^{-6}} = \frac{2.5}{25} \times 10^{3 + 6} = \frac{2.5}{25} \times 10^{3} \]
\[ \beta = 0.1 \times 10^{3} = 100 \]

### Final Answer
The current gain (β) of the transistor is:
\[ \beta = 100 \]

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