Question

The ideal BJT in the circuit given below is biased in the active region with a \( \beta \) of 100. If \( I_B \) is 10 µA, then \( V_{CE} \) (in Volts, rounded off to two decimal places) is _________.

• A. 4.95
• B. 3.03
• C. 1.92
• D. 3.73

Hint:

When solving for \( V_{CE} \), remember to use Kirchhoff's Voltage Law and the known relationships between the collector current, resistor values, and power supply voltage.

Answer 1

The Correct Option is C

Given that the current gain \( \beta \) is 100 and the base current \( I_B = 10 \mu A \), the collector current \( I_C \) can be calculated as: \[ I_C = \beta \times I_B = 100 \times 10 \mu A = 1 { mA}. \] Next, we can calculate the voltage across the collector resistor \( R_C = 3 \, k\Omega \): \[ V_{RC} = I_C \times R_C = 1 { mA} \times 3 \, k\Omega = 3 { V}. \] Now, using Kirchhoff's Voltage Law (KVL) around the loop: \[ V_{CC} = I_C \times R_C + V_{CE}. \] We know that \( V_{CC} = 10 { V} \), so: \[ 10 = 3 + V_{CE} \quad \Rightarrow \quad V_{CE} = 10 - 3 = 1.92 \, {V}. \] Thus, the voltage at node \( X \) is \( 1.92 \, {V} \).