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. 8 k$\Omega$
• B. 4 k$\Omega$
• C. 12 k$\Omega$
• D. 6 k$\Omega$

Hint:

Voltage gain can be conceptualized as $A_v = \text{Current Gain} \times \text{Resistance Gain}$. Here, the current gain is $25$. To reach a target voltage gain of $75$, the circuit needs a resistance gain of exactly $\frac{75}{25} = 3$. This means the output load resistance must be exactly 3 times larger than the input resistance: $3 \times 2\ \text{k}\Omega = 6\ \text{k}\Omega$.

Answer 1

The Correct Option is D

Step 1: Understand what is asked.
In a common-emitter amplifier we are given the current changes, the input resistance, and the voltage gain, and we must find the load resistance $R_L$.
Step 2: Find the current gain $\beta$.
The ac current gain is the ratio of collector-current change to base-current change: $$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{5\ \text{mA}}{0.2\ \text{mA}} = 25.$$
Step 3: Recall the voltage gain formula.
For a CE amplifier, $$A_v = \beta \times \frac{R_L}{R_i}.$$
Step 4: Insert the known quantities.
With $A_v = 75$, $\beta = 25$, and $R_i = 2\ \text{k}\Omega$: $$75 = 25 \times \frac{R_L}{2\ \text{k}\Omega}.$$
Step 5: Simplify the equation.
Dividing both sides by $25$ gives $$3 = \frac{R_L}{2\ \text{k}\Omega}.$$
Step 6: Solve for the load resistance.
Multiplying through by $2\ \text{k}\Omega$: $$R_L = 3 \times 2\ \text{k}\Omega = 6\ \text{k}\Omega.$$
\[ \boxed{R_L = 6\ \text{k}\Omega} \]