Five equal resistances are connected in a network as shown in figure. The net resistance between the points A and B is :

Solution

Physics sir · 4 steps

A is the centre junction and B sits on C. D and E then match, so the DE resistor drops out.

1. Read the wires. A is joined by a plain wire to the centre junction, call it O. B is joined by a plain wire to corner C. So the resistance from A to B is the resistance from O to C.
2. Spot the symmetry. From O, one resistor R goes to D and another goes to E. From D and from E, one resistor R each goes to C. The network looks the same from D as from E, so D and E are at the same potential.
3. Drop the bridge. With no potential difference across it, the resistor between D and E carries no current. This is a balanced Wheatstone bridge, so you can remove that resistor.

   R_(AB) = (R+R)(R+R)/((R+R)+(R+R)) = (2R × 2R)/4R = R

4. So. Two paths O–D–C and O–E–C, each 2R, are in parallel. The net resistance between A and B is R, option (d).

The idea to keep. Follow the bare wires first to find which nodes A and B really are. If two nodes look identical from both ends, they share a potential, and any resistor between them can be removed.

Answer from the NTA final answer key. Solution by PrepWiser.