MHTCET 13 Apr 2026 Shift 2 Paper

Concept:This is a balanced Wheatstone bridge, so the middle branch carries no current.Explanation:For the bridge to be balanced, the ratio of resistances in opposite arms must be equal.Here, $\frac{R_{AB}}{R_{BC}} = \frac{10}{10} = 1$and $\frac{R_{AD}}{R_{DC}} = \frac{30}{30} = 1$.Since both ratios are equal, the bridge is balanced and no current flows through branch $BD$.The circuit then simplifies to two parallel branches between $A$ and $C$.Upper branch resistance: $R_1 = 10 + 10 = 20\ \Omega$Lower branch resistance: $R_2 = 30 + 30 = 60\ \Omega$Equivalent resistance is given by:$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$$$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{60} = \frac{3+1}{60} = \frac{4}{60}$$$$R_{eq} = 15\ \Omega$$Using Ohm's law with battery voltage $V = 3\ \mathrm{V}$:$$I = \frac{V}{R_{eq}} = \frac{3}{15} = 0.2\ \mathrm{A}$$Answer:The current drawn from the battery is $0.2\ \mathrm{A}$ (Option C).