An `N-P-N` transistor is being used in CE mode for which the current transfer ratio is `alpha=(25)/(26)`. The input resistance is `1000Omega` and amplitude of A.C. input voltage is 10 mV. The amplitude of the amplified output collector current is

Options

A. `25muA`
B. `250muA`
C. `125muA`
D. `50muA`

Answer

To solve the problem, we need to find the amplitude of the amplified output collector current (Ic) of an NPN transistor operating in common emitter (CE) mode. Given the parameters are:

- Current transfer ratio (α) = 25/26
- Input resistance (Ri) = 1000 Ω
- Amplitude of AC input voltage (Vi) = 10 mV

### Step-by-Step Solution:

**Step 1: Calculate the Base Current (Ib)**
The base current (Ib) can be calculated using Ohm's law. The formula is:

\[ I_B = \frac{V_I}{R_I} \]

Where:
- \( V_I = 10 \, \text{mV} = 10 \times 10^{-3} \, \text{V} \)
- \( R_I = 1000 \, \Omega \)

Substituting the values:

\[ I_B = \frac{10 \times 10^{-3}}{1000} = 10 \times 10^{-6} \, \text{A} = 10 \, \mu\text{A} \]

**Step 2: Calculate the Current Gain (β)**
The relationship between α and β is given by:

\[ \beta = \frac{\alpha}{1 - \alpha} \]

Substituting the value of α:

\[ \beta = \frac{\frac{25}{26}}{1 - \frac{25}{26}} = \frac{\frac{25}{26}}{\frac{1}{26}} = 25 \]

**Step 3: Calculate the Collector Current (Ic)**
The collector current (Ic) can be calculated using the formula:

\[ I_C = \beta \times I_B \]

Substituting the values of β and Ib:

\[ I_C = 25 \times 10 \times 10^{-6} = 250 \times 10^{-6} \, \text{A} = 250 \, \mu\text{A} \]

### Final Answer:
The amplitude of the amplified output collector current is **250 µA**.

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