A CE amplifier, has a power gain of 40 dB. The input resistance and output load resistance are `500Omega and 2kOmega` respectively. Find the value of common - emitter current gain.
`["take gain in dB "=10log((P_("out"))/(P_("in")))]`

Answer

To solve the problem, we need to find the common-emitter current gain (Ai) of a common-emitter amplifier given its power gain in decibels (dB), input resistance, and output load resistance.

### Step-by-Step Solution:

1. **Convert Power Gain from dB to Linear Scale:**
The power gain in dB is given as:
\[ \text{Gain (dB)} = 10 \log \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) \]
Given that the power gain is 40 dB, we can set up the equation:
\[ 40 = 10 \log \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) \]
Dividing both sides by 10:
\[ 4 = \log \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) \]
Taking the antilogarithm:
\[ \frac{P_{\text{out}}}{P_{\text{in}}} = 10^4 = 10000 \]

2. **Relate Power Gain to Current Gain:**
The power gain (Ap) for a common-emitter amplifier can also be expressed in terms of current gain (Ai):
\[ A_p = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{(I_{\text{out}} \cdot R_{\text{out}})(I_{\text{out}})}{(I_{\text{in}} \cdot R_{\text{in}})(I_{\text{in}})} = \frac{I_{\text{out}}^2 \cdot R_{\text{out}}}{I_{\text{in}}^2 \cdot R_{\text{in}}} \]
Therefore, we can express this as:
\[ A_p = A_i^2 \cdot \frac{R_{\text{out}}}{R_{\text{in}}} \]

3. **Substituting Known Values:**
We know:
- \( A_p = 10000 \)
- \( R_{\text{in}} = 500 \, \Omega \)
- \( R_{\text{out}} = 2000 \, \Omega \)

Substituting these values into the equation:
\[ 10000 = A_i^2 \cdot \frac{2000}{500} \]
Simplifying the fraction:
\[ 10000 = A_i^2 \cdot 4 \]

4. **Solving for Current Gain (Ai):**
Rearranging the equation:
\[ A_i^2 = \frac{10000}{4} = 2500 \]
Taking the square root of both sides:
\[ A_i = \sqrt{2500} = 50 \]

Thus, the common-emitter current gain (Ai) is **50**.