For a transistor `beta = 40` and `I_(B) = 25 muA`. Find the value of `I_(E)`.

Options

A. 1mA
B. `1.025 mA`
C. 2 mA
D. `1.2 mA`

Explanation

To solve the problem, we need to find the emitter current \( I_E \) given the transistor's beta (\( \beta \)) and the base current (\( I_B \)).

### Step-by-Step Solution:

1. **Understand the relationship between the currents:**
The relationship between the collector current (\( I_C \)), base current (\( I_B \)), and emitter current (\( I_E \)) in a transistor is given by:
\[ I_E = I_C + I_B \]

2. **Calculate the collector current (\( I_C \)):**
The beta (\( \beta \)) of the transistor is defined as the ratio of the collector current to the base current:
\[ \beta = \frac{I_C}{I_B} \]
Rearranging this gives:
\[ I_C = \beta \times I_B \]
Given \( \beta = 40 \) and \( I_B = 25 \, \mu A = 25 \times 10^{-6} \, A \), we can substitute these values:
\[ I_C = 40 \times (25 \times 10^{-6}) = 1000 \times 10^{-6} \, A = 1 \, mA \]

3. **Substitute \( I_C \) and \( I_B \) into the emitter current equation:**
Now that we have \( I_C \), we can find \( I_E \):
\[ I_E = I_C + I_B \]
Substituting the values we have:
\[ I_E = 1 \, mA + 25 \, \mu A = 1000 \times 10^{-6} \, A + 25 \times 10^{-6} \, A \]
\[ I_E = (1000 + 25) \times 10^{-6} \, A = 1025 \times 10^{-6} \, A \]
\[ I_E = 1.025 \, mA \]

### Final Answer:
The emitter current \( I_E \) is \( 1.025 \, mA \).

---