What is the power gain in a CE amplifier, where input resistance is `3kOmega` and load resistance 24 `kOmega` given `beta=6` ?

Options

A. 180
B. 288
C. 240
D. 480

Answer

To find the power gain in a common emitter (CE) amplifier given the input resistance, load resistance, and beta, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Input resistance, \( R_i = 3 \, k\Omega = 3000 \, \Omega \)
- Load resistance, \( R_o = 24 \, k\Omega = 24000 \, \Omega \)
- Current gain (beta), \( \beta = 6 \)

2. **Understand the Power Gain Formula:**
The power gain \( P_g \) in a common emitter amplifier can be expressed as:
\[ P_g = \frac{P_{out}}{P_{in}} = \frac{I_C^2 \cdot R_o}{I_B^2 \cdot R_i} \]
where:
- \( I_C \) is the collector current.
- \( I_B \) is the base current.

3. **Relate Collector Current and Base Current:**
From the definition of beta:
\[ \beta = \frac{I_C}{I_B} \implies I_C = \beta \cdot I_B \]

4. **Substitute \( I_C \) in the Power Gain Formula:**
Substitute \( I_C \) in the power gain formula:
\[ P_g = \frac{(\beta \cdot I_B)^2 \cdot R_o}{I_B^2 \cdot R_i} \]
Simplifying this gives:
\[ P_g = \frac{\beta^2 \cdot I_B^2 \cdot R_o}{I_B^2 \cdot R_i} = \beta^2 \cdot \frac{R_o}{R_i} \]

5. **Calculate the Power Gain:**
Now, substituting the values of \( \beta \), \( R_o \), and \( R_i \):
\[ P_g = 6^2 \cdot \frac{24000 \, \Omega}{3000 \, \Omega} \]
\[ P_g = 36 \cdot 8 = 288 \]

6. **Final Answer:**
The power gain in the CE amplifier is:
\[ P_g = 288 \]