Class 12 NEET Physics Current Electricity Questions Part 2

Concept:The four sides of the square loop act as the four arms of a Wheatstone bridge, and the $2\,\Omega$ resistor between $B$ and $D$ is the bridge arm.Solution:The uniform wire has total resistance $4\,\Omega$, so each side of the square has resistance $\frac{4}{4}=1\,\Omega$.Hence $R_{AB}=R_{BC}=R_{CD}=R_{DA}=1\,\Omega$.The battery is connected across $A$ and $C$, and the $2\,\Omega$ resistor is connected between $B$ and $D$.Bridge balance condition: $\frac{R_{AB}}{R_{BC}}=\frac{1}{1}=1$ and $\frac{R_{AD}}{R_{DC}}=\frac{1}{1}=1$.The two ratios are equal, so the bridge is balanced.Equivalently, the circuit is symmetric about the line $AC$, so $V_B=V_D$.Therefore no current flows through the $2\,\Omega$ resistor and it can be removed from the circuit.The circuit between $A$ and $C$ then reduces to two parallel branches:Branch $A\to B\to C$: $1\,\Omega+1\,\Omega=2\,\Omega$.Branch $A\to D\to C$: $1\,\Omega+1\,\Omega=2\,\Omega$.$$R_{\text{eff}}=\frac{2\times 2}{2+2}=1\,\Omega$$Using Ohm's law, $I=\frac{V}{R_{\text{eff}}}=\frac{2\,\text{V}}{1\,\Omega}=2\,\text{A}$.Shortcut:In a balanced Wheatstone bridge, delete the bridge resistor ($2\,\Omega$); the two $2\,\Omega$ arms in parallel give $R_{\text{eff}}=1\,\Omega$, so $I=\frac{2\,\text{V}}{1\,\Omega}=2\,\text{A}$.Answer:$2\,\text{A}$ (Option A)