A transistor has an `alpha = 0.95` It has change in emitter current of 100 mili-ampere, then the change in collector current is :-

Options

A. 95 mA
B. 99.05 mA
C. 0.95 mA
D. 100 mA

Explanation

To solve the problem, we need to find the change in collector current (ΔIc) given the change in emitter current (ΔIe) and the common base current gain (α) of a transistor.

### Step-by-Step Solution:

1. **Understand the relationship between α, ΔIc, and ΔIe**:
The relationship is given by the formula:
\[ \alpha = \frac{\Delta I_c}{\Delta I_e} \]
where:
- α = common base current gain
- ΔIc = change in collector current
- ΔIe = change in emitter current

2. **Given values**:
- α = 0.95
- ΔIe = 100 mA = 100 × 10^{-3} A = 0.1 A

3. **Rearranging the formula to find ΔIc**:
We can rearrange the formula to solve for ΔIc:
\[ \Delta I_c = \alpha \times \Delta I_e \]

4. **Substituting the known values**:
Now, substitute the values of α and ΔIe into the equation:
\[ \Delta I_c = 0.95 \times 0.1 A \]

5. **Calculating ΔIc**:
\[ \Delta I_c = 0.095 A \]
Converting this to milliamperes:
\[ \Delta I_c = 0.095 \times 1000 \, \text{mA} = 95 \, \text{mA} \]

6. **Final Answer**:
The change in collector current (ΔIc) is **95 mA**.