In transistor, the change in base current from 300 microampere to 325 microampere,causes a change in collector current from 7.5 miliampere to 10 miliapmere keeping collector to emitter voltage constant at 10 V.Base to emitter voltage is 0.7V.then the current gain of transistor is ?

Explanation

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

### Step 1: Calculate the change in base current (ΔIb)
The base current changes from 300 microamperes (μA) to 325 microamperes (μA).

\[ \Delta I_B = I_{B(final)} - I_{B(initial)} = 325 \, \mu A - 300 \, \mu A = 25 \, \mu A \]

Convert this to amperes:
\[ \Delta I_B = 25 \, \mu A = 25 \times 10^{-6} \, A \]

### Step 2: Calculate the change in collector current (ΔIc)
The collector current changes from 7.5 milliamperes (mA) to 10 milliamperes (mA).

\[ \Delta I_C = I_{C(final)} - I_{C(initial)} = 10 \, mA - 7.5 \, mA = 2.5 \, mA \]

Convert this to amperes:
\[ \Delta I_C = 2.5 \, mA = 2.5 \times 10^{-3} \, A \]

### Step 3: Calculate the current gain (β)
The current gain (β) of the transistor is given by the formula:

\[ \beta = \frac{\Delta I_C}{\Delta I_B} \]

Substituting the values we calculated:

\[ \beta = \frac{2.5 \times 10^{-3} \, A}{25 \times 10^{-6} \, A} \]

### Step 4: Simplify the calculation
Now, we can simplify the calculation:

\[ \beta = \frac{2.5 \times 10^{-3}}{25 \times 10^{-6}} = \frac{2.5}{25} \times \frac{10^{-3}}{10^{-6}} = 0.1 \times 10^{3} = 100 \]

### Final Answer
Thus, the current gain (β) of the transistor is:

\[ \beta = 100 \]

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