MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Redraw the network using node analysis and simplify parallel and series combinations step by step.Explanation:Label the top junction as $B$, the left lower junction as $A$, and the right lower junction as $C$.The two $3C$ capacitors are both connected between $A$ and $B$, so they are in parallel.Their equivalent is $C_{AB}=3C+3C=6C$.The $1C$ capacitor and the right-side $2C$ capacitor are both between $B$ and $C$, so they are also in parallel.Their equivalent is $C_{BC}=1C+2C=3C$.Now $C_{AB}$ and $C_{BC}$ are connected in series between $A$ and $C$.The series equivalent is $\frac{6C \times 3C}{6C+3C}=2C$.This $2C$ series combination is in parallel with the middle $2C$ capacitor.Thus total equivalent capacitance is $C_{eq}=2C+2C=4C$.Energy stored is $U=\frac{1}{2}C_{eq}V^2$.Substituting $C_{eq}=4C$, we get $U=\frac{1}{2}(4C)V^2=2CV^2$.Answer:Equivalent capacitance is $4C$ and energy stored is $2CV^2$, which matches option A.