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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 resistance of the circuit and current supplied by battery are respectively [Figure: the 12 V battery on the left; R₁ on the top wire leads to a top rail; a bottom rail returns to the battery. Between the top and bottom rails: R₂ vertical, R₃ vertical at the far right, and R₄ on a diagonal wire from the bottom of R₂ up to the top of R₃.]

Asked in JEE Main 5th April 1st Shift 2024 · Series and parallel combinations

• (1) 12 Ω and 1 A
• (2) 10.5 Ω and 1 A
• (3) 12 Ω and 11.4 A
• (4) 10.5 Ω and 1.14 A

Step-by-step solution

Why the other options are wrong

• (2) Pairs a wrong 10.5 Ω with 1 A; these two are not even consistent with 12 V.
• (3) The resistance is right, but no slip gives 11.4 A; it is ten times the 1.14 A of option (d), and 12 V across 12 Ω gives 1 A.
• (4) Uses the sum of conductances 1/8+1/8+1/4=0.5 as the parallel resistance without inverting, giving 10.5 Ω and 12/10.5=1.14 A.

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