The equivalent resistance of the circuit shown below between points a and b is :

Solution

Physics sir · 4 steps

Spot the balanced Wheatstone bridge first: the middle 4 Ω carries no current, so drop it.

1. Find the bridge. Call the top corner T and the centre point C. The path a–T–b has 4 Ω + 4 Ω, the path a–C–b has 4 Ω + 4 Ω, and the vertical 4 Ω joins T to C.
2. Balanced, so remove. The ratios match: 4/4 = 4/4. So T and C are at the same potential, and no current flows through the T–C resistor. You can remove it.
3. Three parallel paths. Now a and b are joined by three separate paths: a–T–b gives 8 Ω, a–C–b gives 8 Ω, and the bottom resistor gives 16 Ω. All three sit between the same two points, so they are in parallel.

   1/R_(ab) = 1/8 + 1/8 + 1/16 = (2 + 2 + 1)/16 = 5/16

4. So. R_(ab) = 16/5 = 3.2 Ω, which is option (c).

The idea to keep. Before writing any equations, check for a Wheatstone bridge. If the arm ratios are equal, the middle resistor carries no current, so delete it and the circuit becomes plain series and parallel.

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