GATE 2025 EC Q.22: solution and answer

Official answer: CSource: Official answer key, GATE 2025 EC (IIT Roorkee)

1. Active region, so $I_C=\beta I_B=100\times 10\mu \text{A}=1\text{mA}$ and $I_E=(\beta +1)I_B=101\times 10\mu \text{A}=1.01\text{mA}$.

2. KCL at the collector node: $R_B$ runs from the collector node to the base, so the current $I_B$ in $R_B$ also leaves this node. The $5\text{k}\Omega$ resistor therefore carries both currents: $I_{5\text{k}}=I_C+I_B=I_E=1.01\text{mA}$.

3. Collector voltage (Ohm's law across the $5\text{k}\Omega$ resistor): $V_C=10-(5\text{k}\Omega )(1.01\text{mA})=10-5.05=4.95\text{V}$.

4. Emitter voltage (Ohm's law across the $3\text{k}\Omega$ resistor, which carries $I_E$): $V_E=(3\text{k}\Omega )(1.01\text{mA})=3.03\text{V}$.

5. Hence $$V_{CE}=V_C-V_E=V_{CC}-I_E(5\text{k}\Omega +3\text{k}\Omega )=10-1.01\times 8=1.92\text{V}$$

6. Check: $V_{CE}=1.92\text{V}$ is well above $V_{CE,\text{sat}}\approx 0.2\text{V}$, and with collector-to-base feedback $V_{CB}=I_B{R}_B>0$, so the collector-base junction is reverse biased and the active-region assumption holds. Neither $R_B$ nor $V_{BE}$ is needed, because $I_B$ is given. Answer: (C).