The input resistance of a common emitter amplifier is `330Omega` and the load resistance is `5 kOmega` A change of base current is `15 muA` results in the change of collector current I mA. The voltage gain of the amplifier is

Options

A. 1000
B. 10001
C. 1010
D. 1100

Answer

To find the voltage gain of a common emitter amplifier, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Input Resistance, \( R_i = 330 \, \Omega \)
- Load Resistance, \( R_L = 5 \, k\Omega = 5000 \, \Omega \)
- Change in Base Current, \( \Delta I_B = 15 \, \mu A = 15 \times 10^{-6} \, A \)
- Change in Collector Current, \( \Delta I_C = 1 \, mA = 1 \times 10^{-3} \, A \)

2. **Calculate Current Gain (\( \beta \)):**
\[ \beta = \frac{\Delta I_C}{\Delta I_B} \]
Substituting the values:
\[ \beta = \frac{1 \times 10^{-3}}{15 \times 10^{-6}} = \frac{1}{15} \times 10^3 = \frac{1000}{15} \approx 66.67 \]

3. **Calculate Voltage Gain (\( V_g \)):**
The voltage gain of a common emitter amplifier is given by:
\[ V_g = \beta \times \frac{R_L}{R_i} \]
Substituting the values:
\[ V_g = 66.67 \times \frac{5000}{330} \]

4. **Calculate \( \frac{R_L}{R_i} \):**
\[ \frac{R_L}{R_i} = \frac{5000}{330} \approx 15.15 \]

5. **Final Calculation of Voltage Gain:**
\[ V_g = 66.67 \times 15.15 \approx 1010.05 \]

### Conclusion:
The voltage gain \( V_g \) of the common emitter amplifier is approximately **1010**.

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