GATE 2026 IN Q.22: solution and answer

Official answer: ASource: Official answer key, GATE 2026 IN (IIT Guwahati)

1. Assume the BJT is in forward-active mode, so $I_C=\beta I_B$ and $I_E=(\beta +1)I_B$, with $V_{BE}=0.65\text{V}$. We check this assumption at the end.

2. Replace the base divider with its Thevenin equivalent: $V_{TH}=5\cdot \frac{5}{5+5}=2.5\text{V}$ and $R_{TH}=5\text{k}\Omega \parallel 5\text{k}\Omega =2.5\text{k}\Omega$.

3. Write KVL around the base-emitter loop, with $R_E=5\text{k}\Omega$: $V_{TH}=I_B{R}_{TH}+V_{BE}+(\beta +1)I_B{R}_E$.

4. Solve for the base current (working in V, k$\Omega$ and mA): $$I_B=\frac{V_{TH}-V_{BE}}{R_{TH}+(\beta +1)R_E}=\frac{2.5-0.65}{2.5+101\times 5}\text{mA}=\frac{1.85}{507.5}\text{mA}\approx 3.645\mu \text{A}$$

5. Collector current: $I_C=\beta I_B=100\times 3.645\mu \text{A}\approx 0.365\text{mA}$.

6. Check the active-mode assumption. $I_E=101I_B\approx 0.368\text{mA}$, so $V_E\approx 0.368\times 5\approx 1.84\text{V}$ and $V_B=V_E+0.65\approx 2.49\text{V}$. Also $V_C=5-0.365\times 5\approx 3.18\text{V}$. Then $V_{CE}\approx 1.34\text{V}$, which is well above $V_{CE(sat)}$, and $V_C>V_B$, so the base-collector junction is reverse biased. The transistor is in forward-active mode, so the assumption holds. The answer is $I_C\approx 0.365\text{mA}$, option (A).