The current I flowing through the given circuit will be __________A.

Solution

Physics sir · 4 steps

Follow the two bare wires and you will find every 9 Ω resistor sits between the same two points.

1. Name the junctions. Call the four dots A, B, C and D from left to right. The battery connects to A and D, and the resistors are A–B, B–C and C–D.
2. Merge joined points. The top wire joins A to C, so A and C are the same point. The bottom wire joins B to D, so B and D are the same point. That leaves only two distinct points, A and B.
3. All in parallel. A–B joins A and B. B–C joins B and A. C–D joins A and B. So all three 9 Ω resistors are connected across A and B, in parallel, and the battery is across A and B too.

   R_(eq) = 9/3 = 3 Ω, I = V/R_(eq) = 6/3

4. So. The current drawn from the battery is I = 2 A, so the answer is 2.

The common slip: 0.22. Treating the three resistors as a plain series chain of 27 Ω gives 6/27 ≈ 0.22 A. The two bare wires make them parallel, not series.

The idea to keep. When wires criss-cross a resistor chain, label each junction and merge every pair joined by a plain wire. Then count which two points each resistor really sits between.

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