Class 12 NEET Physics Current Electricity Questions Part 2

Concept:The square loop of wire forms a balanced Wheatstone bridge, so the $2 \Omega$ resistor carries no current.Formula:For a balanced bridge, the ratio of resistances in opposite arms must be equal.Solution:The total resistance of the uniform wire is $4 \Omega$.Bending it into a square gives each side a resistance of $1 \Omega$.Thus sides $AB$, $BC$, $CD$, and $DA$ are each $1 \Omega$.A $2 \Omega$ resistor is connected between $B$ and $D$ (the bridge).The ratios $AB:BC = 1:1$ and $AD:DC = 1:1$ are equal, so the bridge is balanced.Hence no current flows through the $2 \Omega$ resistor and it can be removed from the circuit.The effective circuit between $A$ and $C$ consists of two parallel paths:Path 1: $A \to B \to C$ with resistance $1 \Omega + 1 \Omega = 2 \Omega$.Path 2: $A \to D \to C$ with resistance $1 \Omega + 1 \Omega = 2 \Omega$.Equivalent resistance $R_{\text{eff}} = \frac{2 \times 2}{2+2} = 1 \Omega$.Using Ohm's law, current $I = \frac{V}{R_{\text{eff}}} = \frac{2\,\text{V}}{1\,\Omega} = 2\,\text{A}$.Shortcut:In a balanced Wheatstone bridge, ignore the bridge resistor ($2 \Omega$).The two arms $ABC$ and $ADC$ are each $2 \Omega$ in parallel, giving $R_{\text{eff}} = 1 \Omega$.Thus $I = 2\,\text{A}$ directly.Answer:$2\,\text{A}$ (Option A)