ISRO EC 2023 Q.14: solution and answer

Official answer: CSource: ISRO's revised and final answer key of 16 February 2024

1. The Q point is the dc operating point $(V_{CE},I_C)$, and it always lies on the dc load line, because $V_{CE}=V_{CC}-I_C{R}_C$ holds for every value of collector current. A change in $\beta$ can therefore only slide the point along the line, never take it off it. Option (a) is ruled out.

2. The question describes the standard textbook case of a transistor biased with a fixed base current, $I_B\approx \frac{V_{CC}-V_{BE}}{R_B}$. This is set by the supply and the base resistor and does not depend on $\beta$.

3. The collector current is $I_C=\beta I_B$, so a larger current gain gives a proportionally larger $I_C$. The Q point does move, which rules out option (b).

4. The larger $I_C$ increases the drop $I_C{R}_C$, so $V_{CE}=V_{CC}-I_C{R}_C$ falls.

5. Higher $I_C$ together with lower $V_{CE}$ is a move toward the saturation end of the load line, which is the upper end when $I_C$ is plotted upward. The Q point moves up on the load line. Option (d) would need $I_C$ to fall, but $I_C=\beta I_B$ rises.