A npn transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance is `100 Omega` and the output load resistance is `10 k Omega`. The common emitter current gain `beta` is:

Options

A. `10^(4)`
B. 60
C. `10^(2)`
D. `6 xx 10^(2)`

Explanation

To solve the problem, we need to find the common emitter current gain (β) of an NPN transistor operating as a common emitter amplifier, given the power gain in decibels, input resistance, and output load resistance.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Power Gain (in dB), \( P_g = 60 \, \text{dB} \)
- Input Resistance, \( R_{in} = 100 \, \Omega \)
- Output Load Resistance, \( R_{out} = 10 \, k\Omega = 10^4 \, \Omega \)

2. **Convert Power Gain from dB to Linear Scale:**
The formula for power gain in decibels is given by:
\[ P_g = 10 \log_{10} \left( \frac{P}{P_0} \right) \]
Rearranging this formula to find the ratio \( \frac{P}{P_0} \):
\[ 60 = 10 \log_{10} \left( \frac{P}{P_0} \right) \]
Dividing both sides by 10:
\[ 6 = \log_{10} \left( \frac{P}{P_0} \right) \]
Converting from logarithmic to linear form:
\[ \frac{P}{P_0} = 10^6 \]

3. **Use the Power Gain Formula:**
The power gain can also be expressed in terms of the current gain (β) and resistances:
\[ P_g = \beta^2 \frac{R_{out}}{R_{in}} \]
Substituting the values we have:
\[ 10^6 = \beta^2 \frac{10^4}{100} \]

4. **Simplify the Equation:**
Simplifying the right side:
\[ \frac{10^4}{100} = 10^2 \]
Therefore, we can rewrite the equation as:
\[ 10^6 = \beta^2 \cdot 10^2 \]

5. **Solve for β:**
Rearranging gives:
\[ \beta^2 = \frac{10^6}{10^2} = 10^4 \]
Taking the square root of both sides:
\[ \beta = 10^2 = 100 \]

6. **Final Answer:**
The common emitter current gain \( \beta \) is:
\[ \beta = 100 \]