If a change of `100 mu A` in the base current of an `n-p-n` transistor in `CE` causes a change of `10 mA` in the collector current, the `ac` current gain of the transistor is

Options

A. 50
B. 100
C. 200
D. 150

Explanation

To find the AC current gain (α) of the transistor, we will use the formula:

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

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

### Step 1: Identify the given values
From the problem, we have:
- Change in base current, \(\Delta I_B = 100 \mu A = 100 \times 10^{-6} A = 10^{-4} A\)
- Change in collector current, \(\Delta I_C = 10 mA = 10 \times 10^{-3} A = 10^{-2} A\)

### Step 2: Substitute the values into the formula
Now we will substitute the values into the formula for α:

\[ \alpha = \frac{\Delta I_C}{\Delta I_B} = \frac{10^{-2}}{10^{-4}} \]

### Step 3: Simplify the expression
Now, we simplify the fraction:

\[ \alpha = \frac{10^{-2}}{10^{-4}} = 10^{-2 + 4} = 10^{2} \]

### Step 4: Calculate the value of α
Now we calculate \(10^{2}\):

\[ \alpha = 100 \]

### Conclusion
Thus, the AC current gain of the transistor is:

\[ \alpha = 100 \]