For a transistor ampliflier in common emiter configuration for load imperdance of `1 k Omega (h_(fe) = 50 and h_(oe) = 25)` the current gain is

Options

A. `-5.2`
B. `-15.7`
C. `-24.8`
D. `-48.76`

Explanation

To find the current gain (Ai) for a transistor amplifier in common emitter configuration, we can follow these steps:

### Step 1: Write down the given values
We are given:
- \( h_{fe} = 50 \)
- \( h_{oe} = 25 \times 10^{-6} \, \text{S} \)
- Load resistance \( R_L = 1 \, \text{k}\Omega = 1000 \, \Omega \)

### Step 2: Use the formula for current gain
The formula for current gain \( A_i \) in a common emitter configuration is given by:

\[ A_i = -\frac{h_{fe}}{1 + h_{oe} \cdot R_L} \]

### Step 3: Substitute the values into the formula
Now, substituting the given values into the formula:

\[ A_i = -\frac{50}{1 + (25 \times 10^{-6}) \cdot (1000)} \]

### Step 4: Calculate the denominator
First, calculate \( h_{oe} \cdot R_L \):

\[ h_{oe} \cdot R_L = 25 \times 10^{-6} \cdot 1000 = 25 \times 10^{-3} = 0.025 \]

Now, substitute this back into the denominator:

\[ 1 + h_{oe} \cdot R_L = 1 + 0.025 = 1.025 \]

### Step 5: Calculate the current gain
Now, substitute this value back into the formula for \( A_i \):

\[ A_i = -\frac{50}{1.025} \]

Calculating this gives:

\[ A_i \approx -48.78 \]

### Step 6: Final answer
Thus, the current gain for the transistor amplifier in common emitter configuration is approximately:

\[ A_i \approx -48.78 \]