Inside a superconducting ring,six identical r | vedclass

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(D) Let the potential at $O$ be $V_O = E$ and the potential at $1$ be $V_1 = 0$. Due to the symmetry of the circuit with respect to the axis passing t

Vedclass · Accepted Answer

Both $(A)$ and $(B)$.

Vedclass · Answer

(D) Let the potential at $O$ be $V_O = E$ and the potential at $1$ be $V_1 = 0$. Due to the symmetry of the circuit with respect to the axis passing through $O$ and $1$,the potentials at points $2$ and $3$ must be equal,i.e.,$V_2 = V_3$. Since $V_2 = V_3$,the potential difference across the resistor $R_{23}$ is $V_2 - V_3 = 0$. Thus,no current flows through $R_{23}$. Furthermore,the circuit is symmetric about the line $O-1$. The resistors $R_{12}$ and $R_{13}$ are connected between the same potential points relative to the symmetry axis,but since $V_2 = V_3$,the current flowing through $R_{12}$ and $R_{13}$ is also zero because there is no potential drop across them relative to the input path. Specifically,the current entering at $O$ splits into three branches: $O-1$,$O-2$,and $O-3$. Since the circuit is symmetric,the current is equally divided into these three resistors. Therefore,statements $(A)$ and $(B)$ are both correct.

Inside a superconducting ring,six identical resistors each of resistance $R$ are connected as shown in the figure. Imagine a battery of $emf$ $E$ connected between points $O$ and $1$,with its positive terminal connected to $O$.

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