The input resistance of a common emitter transistor amplifier, if the output resistance is `500 kOmega`, the current gain `alpha=0.98` and the power gain is `6.0625 xx 10^(6)` is

Options

A. `198 Omega`
B. `300 Omega`
C. `100 Omega`
D. `400 Omega`

Explanation

To find the input resistance of a common emitter transistor amplifier given the output resistance, current gain, and power gain, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Output resistance, \( R_0 = 500 \, k\Omega = 500 \times 10^3 \, \Omega \)
- Current gain, \( \alpha = 0.98 \)
- Power gain, \( P = 6.0625 \times 10^6 \)

2. **Calculate Beta (\( \beta \)):**
- The relationship between \( \alpha \) and \( \beta \) is given by:
\[ \beta = \frac{\alpha}{1 - \alpha} \]
- Substituting the value of \( \alpha \):
\[ \beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49 \]

3. **Relate Power Gain to Voltage Gain and Beta:**
- The power gain \( P \) can be expressed as:
\[ P = A_v \cdot \beta \]
- Rearranging this gives us:
\[ A_v = \frac{P}{\beta} \]
- Substituting the known values:
\[ A_v = \frac{6.0625 \times 10^6}{49} \approx 123.6 \times 10^3 \]

4. **Relate Voltage Gain to Input and Output Resistance:**
- The voltage gain \( A_v \) can also be expressed as:
\[ A_v = \beta \cdot \frac{R_0}{R_i} \]
- Rearranging gives:
\[ R_i = \beta \cdot \frac{R_0}{A_v} \]

5. **Substituting Known Values:**
- Now substituting \( \beta = 49 \), \( R_0 = 500 \times 10^3 \, \Omega \), and \( A_v \):
\[ R_i = 49 \cdot \frac{500 \times 10^3}{123.6 \times 10^3} \]

6. **Calculating Input Resistance:**
- Performing the calculation:
\[ R_i = 49 \cdot \frac{500}{123.6} \approx 198 \, \Omega \]

### Final Answer:
The input resistance \( R_i \) of the common emitter transistor amplifier is approximately **198 ohms**.