A transistor has a current amplification factor (current gain) of 50. In a common emitter amplifier circuit, the collector resistance is chosen as `5Omega` and the input resistance is `1Omega`. The output voltage if input voltage is 0.01 V is

Options

A. -2 V
B. `-5 V`
C. -2.5 V
D. -1 V

Explanation

To find the output voltage \( V_{out} \) in a common emitter amplifier circuit with the given parameters, we can follow these steps:

### Step 1: Understand the relationship between input and output voltages and currents
The current amplification factor (beta, \( \beta \)) is defined as:
\[ \beta = \frac{I_C}{I_B} \]
Where:
- \( I_C \) is the collector current.
- \( I_B \) is the base current.

In terms of voltages and resistances, we can express this as:
\[ \beta = \frac{V_{out}/R_{out}}{V_{in}/R_{in}} \]
Where:
- \( V_{out} \) is the output voltage.
- \( V_{in} \) is the input voltage.
- \( R_{out} \) is the collector resistance.
- \( R_{in} \) is the input resistance.

### Step 2: Substitute the known values into the equation
Given:
- \( \beta = 50 \)
- \( R_{out} = 5 \, \Omega \)
- \( R_{in} = 1 \, \Omega \)
- \( V_{in} = 0.01 \, V \)

We can substitute these values into the equation:
\[ 50 = \frac{V_{out}/5}{0.01/1} \]

### Step 3: Simplify the equation
Rearranging the equation gives:
\[ 50 = \frac{V_{out}}{5} \cdot \frac{1}{0.01} \]
This simplifies to:
\[ 50 = \frac{V_{out}}{5} \cdot 100 \]
Thus:
\[ 50 = \frac{100 \cdot V_{out}}{5} \]
This further simplifies to:
\[ 50 = 20 \cdot V_{out} \]

### Step 4: Solve for \( V_{out} \)
Now, we can solve for \( V_{out} \):
\[ V_{out} = \frac{50}{20} = 2.5 \, V \]

### Conclusion
The output voltage \( V_{out} \) is \( 2.5 \, V \).

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