In a common emitter circuit, if `V_(CE)` is changed by 0.2 V, then collector current changed by `4xx10^(-3)mA`. Output resistance will be-

Options

A. `10kOmega`
B. `30kOmega`
C. `50kOmega`
D. `70kOmega`

Answer

To solve the problem of finding the output resistance in a common emitter circuit given the change in collector current and the change in collector-emitter voltage, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Change in collector-emitter voltage, \( \Delta V_{CE} = 0.2 \, V \)
- Change in collector current, \( \Delta I_C = 4 \times 10^{-3} \, mA = 4 \times 10^{-6} \, A \)

2. **Understand the Relationship for Output Resistance:**
- The output resistance \( R_C \) in a common emitter configuration can be calculated using the formula:
\[ R_C = \frac{\Delta V_{CE}}{\Delta I_C} \]

3. **Substitute the Given Values into the Formula:**
- Plug in the values we have:
\[ R_C = \frac{0.2 \, V}{4 \times 10^{-6} \, A} \]

4. **Calculate the Output Resistance:**
- Performing the division:
\[ R_C = \frac{0.2}{4 \times 10^{-6}} = \frac{0.2}{4} \times 10^{6} = 0.05 \times 10^{6} = 50,000 \, \Omega \]
- Therefore, we can express this in kilo-ohms:
\[ R_C = 50 \, k\Omega \]

5. **Final Answer:**
- The output resistance \( R_C \) is \( 50 \, k\Omega \).