The current transfer ratio `beta` of a transistor is 100. The input resistance of the transistor when used in common emitter mode is 1 kilo ohm. The peak value of the collector alternating current for an input peak voltage of 0.01 volt is

Options

A. `1mA`
B. `10mA`
C. `0.1mA`
D. `0.01mA`

Answer

To find the peak value of the collector alternating current (\(I_C\)) for the given transistor parameters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Current transfer ratio (\(\beta\)) = 100
- Input resistance (\(R_{in}\)) = 1 kΩ = \(1 \times 10^3 \, \Omega\)
- Input peak voltage (\(V_{in}\)) = 0.01 V

2. **Use the Formula for Collector Current**:
The formula to calculate the collector current (\(I_C\)) in a common emitter configuration is:
\[ I_C = \beta \times \frac{V_{in}}{R_{in}} \]

3. **Substitute the Given Values into the Formula**:
\[ I_C = 100 \times \frac{0.01 \, \text{V}}{1 \times 10^3 \, \Omega} \]

4. **Calculate the Value**:
- First, calculate the fraction:
\[ \frac{0.01 \, \text{V}}{1 \times 10^3 \, \Omega} = 0.01 \times 10^{-3} = 10^{-5} \, \text{A} \]
- Now substitute this back into the equation for \(I_C\):
\[ I_C = 100 \times 10^{-5} \, \text{A} = 10^{-3} \, \text{A} \]

5. **Convert the Result to Milliamperes**:
\[ I_C = 10^{-3} \, \text{A} = 1 \, \text{mA} \]

### Final Answer:
The peak value of the collector alternating current (\(I_C\)) is **1 mA**.

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