The input resistance of a common-emitter amplifier is `2 kOmega` and a.c. Current gain is 20. If the load resistance used is `5 kOmega`, calculate the transconductance of the transistor used

Options

A. `0.01 Omega^(-1)`
B. `0.03 Omega^(-1)`
C. `0.04 Omega^(-1)`
D. `0.07 Omega^(-1)`

Answer

To solve the problem, we need to calculate the transconductance (gm) of the transistor used in the common-emitter amplifier. We are given the input resistance (Rin) and the AC current gain (β).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Input Resistance (Rin) = 2 kΩ = 2 × 10³ Ω
- AC Current Gain (β) = 20
- Load Resistance (RL) = 5 kΩ (not directly needed for this calculation)

2. **Formula for Transconductance:**
The transconductance (gm) can be calculated using the formula:
\[ g_m = \frac{\beta}{R_{in}} \]

3. **Substitute the Known Values:**
Substitute the values of β and Rin into the formula:
\[ g_m = \frac{20}{2 \times 10^3} \]

4. **Calculate gm:**
Perform the calculation:
\[ g_m = \frac{20}{2000} = 0.01 \, \text{Ω}^{-1} \]

5. **Final Result:**
Therefore, the transconductance (gm) of the transistor is:
\[ g_m = 0.01 \, \text{Ω}^{-1} \]

### Conclusion:
The correct answer is option 1: **0.01 Ω⁻¹**.