**JEE Advanced 2012 Paper 1, Question 12**

For the resistance network shown in the figure, choose the correct option(s).

- The current through is zero.
- A.
- The potential at is less than that at .
- A.

**Solution**

We can speed up the solution by noticing the mirror symmetry of the circuit. Let the current split into two parts and at the node (see figure). Similarly, let the currents going into node be and . If we were to reverse the polarity of the battery connected to the circuit, all currents would simply reverse direction, but their magnitudes would remain the same. But then, the currents leaving the node must be and since, from the battery’s perspective, the circuit looks the same under . Thus, and .

By similar reasoning the currents through the 1 resistors must be the same, and we label these . Using Kirchoff’s current law we can fill in the remaining currents through and (see figure below).

Applying Kirchoff’s voltage law on the loop ,

Similarly, the loop gives

(1)

Clearly, both of these equations can only be satisfied if . So **option (A) is correct**.

Since no current flows through the 1 resistors and we can treat the legs and as independent. Then,

(2)

and the total current is

(3)

The potential at point is , and that at point is , and so which means **option (C) is correct.**

This, all the given options are correct.