In an `NPN` transistor the collector current is `24 mA`. If `80%` of electrons reach collector it base current in `mA` is

Options

A. 36
B. 26
C. 16
D. 6

Answer

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

### Step-by-Step Solution:

1. **Identify Given Values:**
- Collector current (IC) = 24 mA
- Percentage of electrons reaching the collector = 80%

2. **Relate Emitter Current (IE) to Collector Current (IC):**
- In an NPN transistor, the relationship between the emitter current (IE), collector current (IC), and base current (IB) is given by:
\[ IC = \alpha \times IE \]
- Where α is the common base current gain. In this case, since 80% of the electrons reach the collector, we can say:
\[ IC = 0.8 \times IE \]

3. **Express IE in terms of IC:**
- Rearranging the equation gives:
\[ IE = \frac{IC}{0.8} \]
- Substituting the value of IC:
\[ IE = \frac{24 \, \text{mA}}{0.8} = 30 \, \text{mA} \]

4. **Use the relationship to find Base Current (IB):**
- The relationship between the emitter current, collector current, and base current is:
\[ IE = IC + IB \]
- Rearranging gives:
\[ IB = IE - IC \]
- Substituting the values we found:
\[ IB = 30 \, \text{mA} - 24 \, \text{mA} = 6 \, \text{mA} \]

5. **Final Answer:**
- The base current (IB) is **6 mA**.