MHTCET 11 Apr 2026 Shift 2 Paper

Concept:A balanced Wheatstone bridge carries no current through the middle branch resistor, so the circuit simplifies into two parallel branches.Explanation:The bridge arms are $R_{AB} = 15\ \Omega$, $R_{BC} = 3\ \Omega$, $R_{AD} = 20\ \Omega$, and $R_{DC} = 4\ \Omega$, with a $6\ \Omega$ resistor between $B$ and $D$.Check the balance condition:$$\frac{R_{AB}}{R_{BC}} = \frac{15}{3} = 5$$$$\frac{R_{AD}}{R_{DC}} = \frac{20}{4} = 5$$Since both ratios are equal, the bridge is balanced.Hence, no current flows through the $6\ \Omega$ resistor, so that branch can be removed.The circuit now consists of two parallel branches.Resistance of the upper branch:$$R_1 = 15 + 3 = 18\ \Omega$$Resistance of the lower branch:$$R_2 = 20 + 4 = 24\ \Omega$$The total current supplied is $I = 2.1\ \text{A}$.The $15\ \Omega$ resistor is in the upper branch, so apply the current divider rule:$$I_{15\Omega} = I \times \frac{R_2}{R_1 + R_2}$$Substitute the values:$$I_{15\Omega} = 2.1 \times \frac{24}{18 + 24} = 2.1 \times \frac{24}{42} = 1.2\ \text{A}$$Answer:The current flowing through the $15\ \Omega$ resistance is $1.2\ \text{A}$, which corresponds to option C.