In a n-p-n transistor circuit, the collector current is 10 mA. If 80 percent of electrons emitted reach to the collector. Then the base current (in mA) will be

Answer

To find the base current (I_b) in an n-p-n transistor circuit where the collector current (I_c) is given as 10 mA and 80% of the emitted electrons reach the collector, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between collector current (I_c), emitter current (I_e), and base current (I_b)**:
- In a transistor, the collector current (I_c) is related to the emitter current (I_e) by the equation:
\[ I_c = \alpha I_e \]
- Here, \(\alpha\) is the common base current gain, which is the fraction of the emitter current that reaches the collector. Given that 80% of the electrons emitted reach the collector, we have:
\[ \alpha = 0.8 \]

2. **Express I_e in terms of I_c**:
- Rearranging the equation gives us:
\[ I_e = \frac{I_c}{\alpha} \]
- Substituting the known values:
\[ I_e = \frac{10 \, \text{mA}}{0.8} = 12.5 \, \text{mA} \]

3. **Relate I_e to I_b**:
- The relationship between the emitter current (I_e) and the base current (I_b) can be expressed as:
\[ I_e = I_c + I_b \]
- Rearranging gives:
\[ I_b = I_e - I_c \]

4. **Substitute the values of I_e and I_c**:
- Now substituting the values we found:
\[ I_b = 12.5 \, \text{mA} - 10 \, \text{mA} = 2.5 \, \text{mA} \]

5. **Conclusion**:
- Therefore, the base current (I_b) is:
\[ I_b = 2.5 \, \text{mA} \]

### Final Answer:
The base current (I_b) will be **2.5 mA**.

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