Question

An n-p-n transistor is connected in a circuit as shown in the figure. If \( I_C = 1 \, \text{mA}, \, \beta = 50, \, V_{BE} = 0.7 \, \text{V}, \) and the current through \( R_2 \) is \( 10 I_B \), where \( I_B \) is the base current. Then the ratio \( \frac{R_1}{R_2} \) is: 

• A. 0.375
• B. 0.25
• C. 0.5
• D. 0.275

Hint:

In transistor circuits, base, collector, and emitter currents are related by \( I_C = \beta I_B \) and Ohm's law is used to calculate voltage drops across resistors.

Answer 1

The Correct Option is B

Step 1: Understanding the transistor current relations.
The current through \( R_2 \) is given by \( I_{R_2} = 10 I_B \). The collector current \( I_C \) is related to the base current by \( I_C = \beta I_B \), so \( I_B = \frac{I_C}{\beta} = \frac{1 \, \text{mA}}{50} = 20 \, \mu\text{A} \).
Step 2: Calculating the voltage drop across \( R_2 \).
Using Ohm's law, the voltage drop across \( R_2 \) is \( V_{R_2} = I_{R_2} R_2 = 10 I_B R_2 = 10 \times 20 \, \mu\text{A} \times R_2 \).
Step 3: Calculating the voltage at the collector.
The voltage at the collector is \( V_C = 6 \, \text{V} - V_{R_2} = 3 \, \text{V} \), so solving for \( R_2 \) and then \( R_1 \), we find the ratio \( \frac{R_1}{R_2} = 0.25 \).
Step 4: Conclusion.
Thus, the correct answer is option (B).