MHTCET 17 Apr 2026 Shift 1 Paper

Concept:By Kirchhoff's Current Law, the total current entering a junction is exactly equal to the total current leaving that junction. This principle is used step by step to find the unknown current $I$.Explanation:Let the three junctions from left to right be labelled as $J_1$, $J_2$, and $J_3$.At junction $J_1$, the currents entering are $4.5\ \mathrm{A}$ and $1.7\ \mathrm{A}$, while the currents leaving are $2.3\ \mathrm{A}$ and $I_1$.Applying KCL at $J_1$:$$4.5 + 1.7 = 2.3 + I_1$$$$I_1 = 6.2 - 2.3 = 3.9\ \mathrm{A}$$So, the current flowing from $J_1$ to $J_2$ is $3.9\ \mathrm{A}$.At junction $J_2$, the entering currents are $I_1$, $3.2\ \mathrm{A}$, and $4.9\ \mathrm{A}$, while the leaving currents are $2.6\ \mathrm{A}$ and $I_2$.Applying KCL at $J_2$:$$I_1 + 3.2 + 4.9 = 2.6 + I_2$$$$3.9 + 3.2 + 4.9 = 12.0 = 2.6 + I_2$$$$I_2 = 12.0 - 2.6 = 9.4\ \mathrm{A}$$Thus, the current reaching junction $J_3$ is $9.4\ \mathrm{A}$.At junction $J_3$, the entering current is $I_2$, and the leaving currents are $1.9\ \mathrm{A}$ and $I$.Applying KCL at $J_3$:$$I_2 = 1.9 + I$$$$9.4 = 1.9 + I$$$$I = 9.4 - 1.9 = 7.5\ \mathrm{A}$$Hence, the required current $I$ is obtained.Answer:$I = 7.5\ \mathrm{A}$, so the correct option is D.