In a common emitter transistor amplifier, the output resistance is `500KOmega` and the current gain `beta=49`. If the power gain of the amplifier is `5xx10^(6)`, the input resistance is

Options

A. `165Omega`
B. `198Omega`
C. `225 Omega`
D. `240Omega`

Answer

To find the input resistance \( R_i \) of the common emitter transistor amplifier, we can use the relationship between power gain, current gain, output resistance, and input resistance.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Output resistance \( R_0 = 500 \, \text{k}\Omega = 500 \times 10^3 \, \Omega \)
- Current gain \( \beta = 49 \)
- Power gain \( P_g = 5 \times 10^6 \)

2. **Use the Power Gain Formula:**
The power gain \( P_g \) can be expressed in terms of current gain \( \beta \) and resistances:
\[ P_g = \beta^2 \times \frac{R_0}{R_i} \]

3. **Rearrange the Formula to Solve for Input Resistance \( R_i \):**
Rearranging the equation gives:
\[ R_i = \beta^2 \times \frac{R_0}{P_g} \]

4. **Calculate \( \beta^2 \):**
\[ \beta^2 = 49^2 = 2401 \]

5. **Substitute the Values into the Equation:**
Now substitute \( \beta^2 \), \( R_0 \), and \( P_g \) into the equation:
\[ R_i = 2401 \times \frac{500 \times 10^3}{5 \times 10^6} \]

6. **Perform the Calculation:**
First, calculate \( \frac{500 \times 10^3}{5 \times 10^6} \):
\[ \frac{500 \times 10^3}{5 \times 10^6} = \frac{500}{5} \times \frac{10^3}{10^6} = 100 \times 10^{-3} = 0.1 \]
Now substitute back:
\[ R_i = 2401 \times 0.1 = 240.1 \, \Omega \]

7. **Final Result:**
Rounding off, we get:
\[ R_i \approx 240 \, \Omega \]

### Conclusion:
The input resistance \( R_i \) of the amplifier is approximately \( 240 \, \Omega \).