Question

In common emitter amplifier, a change of \(0.2\) mA in the base current causes a change of \(5\) mA in the collector current. If input resistance is \(2\) k\(\Omega\) and voltage gain is \(75\), the load resistance used in the circuit is

• A. \(2\) k\(\Omega\)
• B. \(3\) k\(\Omega\)
• C. \(4\) k\(\Omega\)
• D. \(6\) k\(\Omega\)

Hint:

Voltage gain = beta x (R_L / R_in), with beta = delta Ic / delta Ib.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The current gain is \(\beta = \dfrac{\Delta I_C}{\Delta I_B}\), and the voltage gain of a CE amplifier is \(A_V = \beta\dfrac{R_L}{R_{in}}\).

Step 2: Compute
\[ \beta = \frac{5 \text{ mA}}{0.2 \text{ mA}} = 25 \]
\[ 75 = 25 \times \frac{R_L}{2\text{ k}\Omega} \Rightarrow R_L = \frac{75 \times 2}{25} = 6\text{ k}\Omega \]
Option (B), 3 kilo-ohm, would give a voltage gain of only 37.5.

Final Answer:
The load resistance is 6 k\(\Omega\), option (D). \[ \boxed{6\text{ k}\Omega} \]