In a common emitter amplifier, using output reisistance of `5000` ohm and input resistance fo `2000`ohm, if the peak value of input signal voltage is `10 m V` and `beta=50`, then the peak value of output voltage is

Options

A. `5xx10^(-6)V`
B. `2.5xx10^(-4)V`
C. `1.25V`
D. `125V`

Answer

To solve the problem step by step, we will use the given parameters of the common emitter amplifier.

### Step 1: Identify the given values
- Output resistance \( R_O = 5000 \, \Omega \)
- Input resistance \( R_I = 2000 \, \Omega \)
- Peak input voltage \( V_I = 10 \, \text{mV} = 10 \times 10^{-3} \, \text{V} \)
- Current gain \( \beta = 50 \)

### Step 2: Calculate the input current \( I_B \)
Using Ohm's Law, the input current \( I_B \) can be calculated as:
\[ I_B = \frac{V_I}{R_I} \]
Substituting the values:
\[ I_B = \frac{10 \times 10^{-3}}{2000} = 5 \times 10^{-6} \, \text{A} = 5 \, \mu\text{A} \]

### Step 3: Calculate the collector current \( I_C \)
The relationship between the collector current \( I_C \) and the base current \( I_B \) is given by:
\[ I_C = \beta \times I_B \]
Substituting the values:
\[ I_C = 50 \times 5 \times 10^{-6} = 250 \times 10^{-6} \, \text{A} = 0.25 \, \text{mA} \]

### Step 4: Calculate the output voltage \( V_O \)
The output voltage \( V_O \) can be calculated using the formula:
\[ V_O = I_C \times R_O \]
Substituting the values:
\[ V_O = 0.25 \times 10^{-3} \times 5000 \]
Calculating this gives:
\[ V_O = 1.25 \, \text{V} \]

### Final Answer
The peak value of the output voltage \( V_O \) is \( 1.25 \, \text{V} \).