In a common emitter transistor circuit, the base current is `40 muA` , then `V_(BE)` is

Options

A. `2 V`
B. `0.2 V`
C. `0.8 V`
D. Zero

Answer

To find the base-emitter voltage \( V_{BE} \) in a common emitter transistor circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Base current \( I_B = 40 \, \mu A = 40 \times 10^{-6} A \)
- Resistor connected to the base \( R_B = 245 \, k\Omega = 245 \times 10^{3} \Omega \)
- Supply voltage \( V_{CC} = 10 \, V \)

2. **Use Kirchhoff’s Voltage Law:**
According to Kirchhoff's Voltage Law, we can write the equation for the loop involving \( V_{CC} \), the voltage drop across \( R_B \), and \( V_{BE} \):
\[ V_{CC} - I_B \cdot R_B - V_{BE} = 0 \]

3. **Rearranging the Equation:**
Rearranging the equation to solve for \( V_{BE} \):
\[ V_{BE} = V_{CC} - I_B \cdot R_B \]

4. **Substituting the Values:**
Substitute the known values into the equation:
\[ V_{BE} = 10 \, V - (40 \times 10^{-6} A) \cdot (245 \times 10^{3} \Omega) \]

5. **Calculating the Voltage Drop Across \( R_B \):**
Calculate \( I_B \cdot R_B \):
\[ I_B \cdot R_B = 40 \times 10^{-6} \cdot 245 \times 10^{3} = 9.8 \, V \]

6. **Final Calculation of \( V_{BE} \):**
Now substitute back into the equation for \( V_{BE} \):
\[ V_{BE} = 10 \, V - 9.8 \, V = 0.2 \, V \]

7. **Conclusion:**
Therefore, the value of \( V_{BE} \) is \( 0.2 \, V \).

### Final Answer:
\[ V_{BE} = 0.2 \, V \]