The equivalent resistance between A and B of the network shown in figure;

Solution

Physics sir · 3 steps

Spot the Wheatstone bridge: check the ratios, and if they match, the 9R in the middle carries no current.

1. Name the nodes. Call the junction after 2R point C and the junction before 3R point D. Then 2R joins A–C, R joins A–D, 9R joins C–D, 6R joins C–B and 3R joins D–B. This is a Wheatstone bridge with 9R as the middle arm.
2. Check balance. On the upper path the ratio is R/3R = 1/3. On the lower path it is 2R/6R = 1/3. The ratios are equal, so C and D are at the same potential and no current flows through 9R. You can remove it.

   R_(AB) = (2R + 6R)(R + 3R)/((2R + 6R) + (R + 3R)) = (8R × 4R)/12R = 8/3R

3. So. With 9R gone you have two branches, 8R and 4R, in parallel. The equivalent resistance is 8/3R, option (b).

The idea to keep. When five resistors form a diamond with one across the middle, check the ratios first. If they match, the bridge is balanced: drop the middle resistor and you are left with two series branches in parallel.

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