A transistor connected in common emitter configuration has input resistance `R_("in")=2 K Omega` and load resistance of `5 K Omega` . If `beta=60` and an input signal 12 mV is applied , calculate the resistance gain, voltage gain and power gain.

Options

A. 9000
B. 4000
C. 6000
D. 8000

Answer

To solve the problem step by step, we will calculate the resistance gain, voltage gain, and power gain for the given transistor in common emitter configuration.

### Step 1: Calculate Resistance Gain
The resistance gain (R_gain) is defined as the ratio of the output resistance (load resistance, \( R_L \)) to the input resistance (\( R_{in} \)).

Given:
- Input resistance, \( R_{in} = 2 \, k\Omega = 2000 \, \Omega \)
- Load resistance, \( R_L = 5 \, k\Omega = 5000 \, \Omega \)

Using the formula:
\[ R_{gain} = \frac{R_L}{R_{in}} = \frac{5000 \, \Omega}{2000 \, \Omega} = 2.5 \]

### Step 2: Calculate Voltage Gain
The voltage gain (V_gain) can be calculated using the formula:
\[ V_{gain} = \beta \times R_{gain} \]

Given:
- Current amplification factor, \( \beta = 60 \)

Substituting the values:
\[ V_{gain} = 60 \times 2.5 = 150 \]

### Step 3: Calculate Power Gain
The power gain (P_gain) is given by the formula:
\[ P_{gain} = \beta^2 \times \frac{R_L}{R_{in}} \]

Substituting the values:
\[ P_{gain} = 60^2 \times \frac{5000 \, \Omega}{2000 \, \Omega} \]
\[ P_{gain} = 3600 \times 2.5 = 9000 \]

### Final Answers:
- Resistance Gain: \( 2.5 \)
- Voltage Gain: \( 150 \)
- Power Gain: \( 9000 \)