In an `NPN` transistor the collector current is `10 mA`. If `90%` of electrons reach collector, the emitter current `(i_(E))` and base current `(i_(B))` are given by

Options

A. `I_(E) = 1mA ,I_(B) = 11mA`
B. `I_(E) = 11mA , I_(B) = 1mA`
C. `I_(E) = 1mA, I_(B) = 9mA`
D. `I_(E) = 9mA , I_(B) = -1 mA`

Answer

To solve the problem, we need to determine the emitter current \( I_E \) and the base current \( I_B \) for the given NPN transistor, where the collector current \( I_C \) is \( 10 \, \text{mA} \) and \( 90\% \) of the electrons reach the collector.

### Step-by-Step Solution:

1. **Identify the given values**:
- Collector current \( I_C = 10 \, \text{mA} \)
- Percentage of electrons reaching the collector = \( 90\% \)

2. **Relate the collector current to the emitter current using the common base current gain (α)**:
- The relationship between collector current \( I_C \), emitter current \( I_E \), and α is given by:
\[ \alpha = \frac{I_C}{I_E} \]
- Since \( 90\% \) of the electrons reach the collector, we can express α as:
\[ \alpha = 0.9 \]

3. **Calculate the emitter current \( I_E \)**:
- Rearranging the formula for α gives:
\[ I_E = \frac{I_C}{\alpha} \]
- Substituting the known values:
\[ I_E = \frac{10 \, \text{mA}}{0.9} \approx 11.11 \, \text{mA} \]

4. **Calculate the base current \( I_B \)**:
- The base current can be calculated using the relationship:
\[ I_E = I_C + I_B \]
- Rearranging gives:
\[ I_B = I_E - I_C \]
- Substituting the values we found:
\[ I_B = 11.11 \, \text{mA} - 10 \, \text{mA} \approx 1.11 \, \text{mA} \]

5. **Final Results**:
- Emitter current \( I_E \approx 11.11 \, \text{mA} \)
- Base current \( I_B \approx 1.11 \, \text{mA} \)

### Summary of Results:
- Emitter current \( I_E \approx 11.11 \, \text{mA} \)
- Base current \( I_B \approx 1.11 \, \text{mA} \)