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 solve the problem of finding the voltage gain of a common emitter amplifier, we can follow these steps:

### Step 1: Understand the Parameters
We are given:
- Input resistance, \( R_{in} = 330 \, \Omega \)
- 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, \( \Delta I_C = 1 \, mA = 1 \times 10^{-3} \, A \)

### Step 2: Calculate the Current Gain
The current gain \( \beta \) (or \( h_{fe} \)) of the amplifier can be calculated using the formula:
\[ \beta = \frac{\Delta I_C}{\Delta I_B} \]
Substituting the values:
\[ \beta = \frac{1 \times 10^{-3}}{15 \times 10^{-6}} = \frac{1}{15} \times 10^{3} = \frac{1000}{15} \approx 66.67 \]

### Step 3: Calculate the Voltage Gain
The voltage gain \( A_v \) of the common emitter amplifier can be calculated using the formula:
\[ A_v = \beta \times \frac{R_L}{R_{in}} \]
Substituting the values we have:
\[ A_v = 66.67 \times \frac{5000}{330} \]

### Step 4: Simplify the Expression
First, calculate \( \frac{5000}{330} \):
\[ \frac{5000}{330} \approx 15.15 \]
Now substitute this back into the voltage gain formula:
\[ A_v \approx 66.67 \times 15.15 \approx 1010.05 \]

### Step 5: Final Result
Thus, the voltage gain of the amplifier is approximately:
\[ A_v \approx 1010 \]

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

---