A transsitor is operated in CE configuration at `V_(CC) = 2 V` such that a change in base current from 100 `muA` to 200 `muA` produces a change in the collector current from 9mA to 16.5 mA. The value of current gain, `beta` is -

Options

A. 45
B. 50
C. 60
D. 75

Answer

To find the current gain (β) of the transistor in the common emitter (CE) configuration, we can use the formula:

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

where:
- \(\Delta I_C\) is the change in collector current,
- \(\Delta I_B\) is the change in base current.

### Step 1: Calculate the change in collector current (\(\Delta I_C\))

Given:
- Initial collector current, \(I_{C1} = 9 \, \text{mA} = 9 \times 10^{-3} \, \text{A}\)
- Final collector current, \(I_{C2} = 16.5 \, \text{mA} = 16.5 \times 10^{-3} \, \text{A}\)

Now, we calculate \(\Delta I_C\):

\[ \Delta I_C = I_{C2} - I_{C1} = (16.5 \times 10^{-3}) - (9 \times 10^{-3}) = 7.5 \times 10^{-3} \, \text{A} \]

### Step 2: Calculate the change in base current (\(\Delta I_B\))

Given:
- Initial base current, \(I_{B1} = 100 \, \mu A = 100 \times 10^{-6} \, \text{A}\)
- Final base current, \(I_{B2} = 200 \, \mu A = 200 \times 10^{-6} \, \text{A}\)

Now, we calculate \(\Delta I_B\):

\[ \Delta I_B = I_{B2} - I_{B1} = (200 \times 10^{-6}) - (100 \times 10^{-6}) = 100 \times 10^{-6} \, \text{A} \]

### Step 3: Substitute the values into the formula for β

Now we can substitute \(\Delta I_C\) and \(\Delta I_B\) into the formula for β:

\[ \beta = \frac{\Delta I_C}{\Delta I_B} = \frac{7.5 \times 10^{-3}}{100 \times 10^{-6}} \]

### Step 4: Simplify the expression

Calculating the above expression:

\[ \beta = \frac{7.5 \times 10^{-3}}{100 \times 10^{-6}} = \frac{7.5}{100} \times 10^{3} = 0.075 \times 10^{3} = 75 \]

### Conclusion

Thus, the value of the current gain \(\beta\) is:

\[ \beta = 75 \]

### Final Answer

The value of current gain, \(\beta\) is **75**.

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