The dc common emitter current gain of a n-p-n transistor is 50. The potential difference applied across the collector and emitter of a transistor used in CE configuration is, `V_(CE)=2V`. If the collector resistance, `R_(C)=4kOmega,` the base current `(I_B)` and the collector current `(I_(C))` are

Options

A. `I_(B)=10 muA,I_(C)=0.5mA`
B. `I_(B)=0.5 muA,I_(C)=10mA`
C. `I_(B)=5 muA,I_(C)=1mA`
D. `I_(B)=1 muA,I_(C)=0.5mA`

Answer

To solve the problem, we need to find the base current \( I_B \) and the collector current \( I_C \) for the given NPN transistor in common emitter configuration.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Common emitter current gain (β) = 50
- Collector-emitter voltage (\( V_{CE} \)) = 2 V
- Collector resistance (\( R_C \)) = 4 kΩ = 4000 Ω

2. **Calculate the Collector Current (\( I_C \)):**
The collector current can be calculated using Ohm's law:
\[ I_C = \frac{V_{CE}}{R_C} \]
Substituting the values:
\[ I_C = \frac{2 \, \text{V}}{4000 \, \Omega} = \frac{2}{4000} = 0.0005 \, \text{A} = 0.5 \, \text{mA} \]

3. **Use the Current Gain to Find Base Current (\( I_B \)):**
The relationship between collector current, base current, and current gain is given by:
\[ \beta = \frac{I_C}{I_B} \]
Rearranging this gives:
\[ I_B = \frac{I_C}{\beta} \]
Substituting the values we have:
\[ I_B = \frac{0.5 \, \text{mA}}{50} = \frac{0.5 \times 10^{-3} \, \text{A}}{50} = 0.01 \, \text{mA} = 10 \, \mu\text{A} \]

### Final Results:
- Collector Current \( I_C = 0.5 \, \text{mA} \)
- Base Current \( I_B = 10 \, \mu\text{A} \)