The transfer ratio of a transistor is `50`. The input resistance of the transistor when used in the common -emitter configuration is `1 kOmega`. The peak value for an `A.C.` input voltage of `0.01 V` peak is

Options

A. `100 muA`
B. `0.01 muA`
C. `0.25 mu A`
D. `500 muA`

Answer

To solve the problem step by step, we will use the given values and apply the relevant formulas.

### Step 1: Identify the given values
- Transfer ratio (current gain, β) = 50
- Input resistance (R_input) = 1 kΩ = 1000 Ω
- Peak value of AC input voltage (V_in) = 0.01 V

### Step 2: Calculate the base current (I_B)
The base current can be calculated using Ohm's law:
\[ I_B = \frac{V_{in}}{R_{input}} \]
Substituting the values:
\[ I_B = \frac{0.01 \, \text{V}}{1000 \, \Omega} = \frac{0.01}{1000} = 0.00001 \, \text{A} = 10^{-5} \, \text{A} = 10 \, \mu A \]

### Step 3: Calculate the collector current (I_C)
The collector current can be calculated using the formula:
\[ I_C = \beta \times I_B \]
Substituting the values:
\[ I_C = 50 \times 10^{-5} \, \text{A} = 5 \times 10^{-4} \, \text{A} \]
This can also be expressed in microamperes:
\[ I_C = 500 \, \mu A \]

### Final Result
The peak value for the collector current (I_C) is:
\[ I_C = 500 \, \mu A \]