To solve the problem step by step, we will calculate the voltage gain and power gain of the given amplifier with the provided parameters.
### Step 1: Calculate the Load Current (I_L)
The load current \( I_L \) can be calculated using the formula:
\[
I_L = \frac{V_L}{R_L}
\]
where:
- \( V_L = 0.8 \, V \) (voltage drop across the load resistance)
- \( R_L = 800 \, \Omega \) (load resistance)
Substituting the values:
\[
I_L = \frac{0.8 \, V}{800 \, \Omega} = 0.001 \, A = 1 \, mA
\]
### Step 2: Calculate the Collector Current (I_C)
The current amplification factor \( \alpha \) is given as 0.96. The relationship between the collector current \( I_C \) and the base current \( I_B \) is given by:
\[
\alpha = \frac{I_C}{I_E} \quad \text{(where } I_E \text{ is the emitter current)}
\]
Since \( I_E \approx I_C \) for large \( \beta \), we can express \( I_B \) in terms of \( I_C \):
\[
I_B = \frac{I_C}{\alpha}
\]
From the previous step, we know that \( I_L = I_C \):
\[
I_B = \frac{I_C}{0.96} = \frac{1 \, mA}{0.96} \approx 1.04167 \, mA \approx \frac{25}{24} \, mA
\]
### Step 3: Calculate the Input Voltage (V_input)
The input voltage \( V_{input} \) can be calculated using:
\[
V_{input} = I_B \times R_I
\]
where:
- \( R_I = 192 \, \Omega \) (input resistance)
Substituting the values:
\[
V_{input} = \left( \frac{25}{24} \times 10^{-3} \, A \right) \times 192 \, \Omega
\]
Calculating \( V_{input} \):
\[
V_{input} = \frac{25 \times 192}{24} \times 10^{-3} = 200 \times 10^{-3} = 0.2 \, V
\]
### Step 4: Calculate the Voltage Gain (A_V)
The voltage gain \( A_V \) is given by:
\[
A_V = \frac{V_{load}}{V_{input}}
\]
Substituting the values:
\[
A_V = \frac{0.8 \, V}{0.2 \, V} = 4
\]
### Step 5: Calculate the Power Gain (A_P)
The power gain \( A_P \) can be calculated using:
\[
A_P = \frac{I_C^2 \times R_L}{I_B^2 \times R_I}
\]
Substituting the values:
\[
A_P = \frac{(1 \times 10^{-3})^2 \times 800}{\left(\frac{25}{24} \times 10^{-3}\right)^2 \times 192}
\]
Calculating \( A_P \):
\[
A_P = \frac{(1 \times 10^{-6}) \times 800}{\left(\frac{625}{576} \times 10^{-6}\right) \times 192}
\]
\[
= \frac{800 \times 10^{-6}}{\frac{625 \times 192}{576} \times 10^{-6}} = \frac{800 \times 576}{625 \times 192} = \frac{460800}{120000} \approx 3.84
\]
### Final Results
Thus, the voltage gain \( A_V \) is 4 and the power gain \( A_P \) is approximately 3.84.
### Summary
- Voltage Gain \( A_V = 4 \)
- Power Gain \( A_P \approx 3.84 \)
