The total current supplied to the circuit as shown in figure by the 5 V battery is ____________ A.

Solution

Physics sir · 4 steps

Both slanted 5 Ω resistors start on the same top wire, so they are in parallel, not in a Y.

1. Spot the nodes. The whole top wire is one node at the battery's positive end, and the whole bottom wire is one node at its negative end. The two slanted 5 Ω resistors both run from the top wire to the same middle junction, so they sit across the same two points.
2. Middle branch. Two 5 Ω in parallel give 2.5 Ω. This is in series with the 2.5 Ω going down to the bottom wire, so the middle branch is 2.5 + 2.5 = 5 Ω.
3. Two equal branches. The right-hand 5 Ω also connects the top wire to the bottom wire. So the battery sees two 5 Ω branches in parallel, which is 2.5 Ω in total.

   I = V/R_(eq) = (5 V)/(2.5 Ω) = 2 A

4. So. Each branch carries 1 A, and the battery supplies 2 A in total.

The common slip: 1. Leaving out the right-hand 5 Ω branch gives only the middle branch's current, 1 A, not the total of 2 A.

The idea to keep. Before using Kirchhoff's laws, trace the plain wires: every point on one unbroken wire is the same node. Resistors that join the same two nodes are in parallel, whatever their shape on paper.

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