For a common emitter transistor working in active state, following data is given `R_(L)=1KOmega` `V_("in")=10mV`
`triangleI_(B)=15muA,triangleI_(C)=3mA`. The input resistance `r_(i)`& voltage gain `A_(V)` for the transitor is

Options

A. `200,0.67KOmega`
B. `300,0.67KOmega`
C. `200,0.1KOmega`
D. `300,1KOmega`

Answer

To solve the problem, we will find the input resistance \( r_i \) and the voltage gain \( A_V \) for the common emitter transistor using the given data.

### Step 1: Calculate Input Resistance \( r_i \)

The formula for the input resistance \( r_i \) is given by:

\[ r_i = \frac{V_{in}}{\Delta I_B} \]

Where:
- \( V_{in} = 10 \, \text{mV} = 10 \times 10^{-3} \, \text{V} \)
- \( \Delta I_B = 15 \, \mu A = 15 \times 10^{-6} \, A \)

Substituting the values into the formula:

\[ r_i = \frac{10 \times 10^{-3}}{15 \times 10^{-6}} = \frac{10}{15} \times 10^3 \, \Omega = \frac{2}{3} \times 10^3 \, \Omega \approx 0.67 \, k\Omega \]

### Step 2: Calculate the Current Gain \( \beta \)

The current gain \( \beta \) is given by:

\[ \beta = \frac{\Delta I_C}{\Delta I_B} \]

Where:
- \( \Delta I_C = 3 \, mA = 3 \times 10^{-3} \, A \)

Substituting the values:

\[ \beta = \frac{3 \times 10^{-3}}{15 \times 10^{-6}} = \frac{3}{15} \times 10^3 = \frac{1}{5} \times 10^3 = 200 \]

### Step 3: Calculate Voltage Gain \( A_V \)

The formula for the voltage gain \( A_V \) is given by:

\[ A_V = -\frac{\beta \cdot R_L}{r_i} \]

Where:
- \( R_L = 1 \, k\Omega = 1000 \, \Omega \)

Substituting the values:

\[ A_V = -\frac{200 \cdot 1000}{0.67 \times 10^3} \]

Calculating the denominator:

\[ A_V = -\frac{200000}{670} \approx -298.51 \]

Rounding this, we find:

\[ A_V \approx -300 \]

### Final Answers

- Input Resistance \( r_i \approx 0.67 \, k\Omega \)
- Voltage Gain \( A_V \approx -300 \)