In the given figure R₁=10 Ω, R₂=8 Ω, R₃=4 Ω and R₄=8 Ω. Battery is ideal with emf 12 V. Equivalent resistant of the circuit and current supplied by battery are respectively :

Solution

Physics sir · 4 steps

Follow the wires: R2, R4 and R3 all join the same two nodes, so they are in parallel.

1. Find the nodes. The top ends of R₂, R₄ and R₃ all touch the top wire after R₁. Their bottom ends all touch the bottom wire, which leads back to the battery. All three sit between the same two nodes, so they are in parallel.

   1/Rₚ = 1/8 + 1/8 + 1/4 = 1/2 ⇒ Rₚ = 2 Ω

2. Add R1 in series. All the battery current passes through R₁ before it splits, so R₁ is in series with this group. R_(eq) = 10 + 2 = 12 Ω.
3. Battery current. For an ideal battery, I = E/R_(eq) = (12 V)/(12 Ω) = 1 A.
4. So. The equivalent resistance is 12 Ω and the battery supplies 1 A, which is option (d).

The idea to keep. Do not judge a connection by how slanted a resistor is drawn. Mark the nodes, and resistors that share the same two nodes are in parallel, whatever the picture looks like.

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