In n-p-n transistor circuit, the collector current is `10 mA`. If `90%` of the holes reach the collector, find emitter and base currents.

Options

A. 10 mA, 1 mA
B. 22 mA,11 mA
C. 11 mA, 1 mA
D. 20 mA, 10 mA

Explanation

To solve the problem, we need to find the emitter current (IE) and the base current (IB) of the NPN transistor given the collector current (IC) and the percentage of holes reaching the collector.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Collector Current, \( I_C = 10 \, \text{mA} \)
- Percentage of holes reaching the collector = 90%

2. **Relation Between Collector Current and Emitter Current:**
- Since 90% of the emitted holes reach the collector, we can express this relationship mathematically:
\[ I_C = 0.9 \times I_E \]

3. **Rearranging the Equation to Find Emitter Current (IE):**
- To find \( I_E \), we rearrange the equation:
\[ I_E = \frac{I_C}{0.9} \]

4. **Substituting the Value of IC:**
- Now substitute \( I_C = 10 \, \text{mA} \) into the equation:
\[ I_E = \frac{10 \, \text{mA}}{0.9} \approx 11.11 \, \text{mA} \]

5. **Calculate Base Current (IB):**
- The relationship between emitter current, collector current, and base current is given by:
\[ I_E = I_B + I_C \]
- Rearranging gives:
\[ I_B = I_E - I_C \]

6. **Substituting the Values of IE and IC:**
- Now substitute the values we have:
\[ I_B = 11.11 \, \text{mA} - 10 \, \text{mA} = 1.11 \, \text{mA} \]

### Final Results:
- Emitter Current, \( I_E \approx 11.11 \, \text{mA} \)
- Base Current, \( I_B \approx 1.11 \, \text{mA} \)