In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: R₁=15 Ω, R₂=10 Ω, R₃=20 Ω, R₄=5 Ω, R₅=25 Ω, R₆=30 Ω, E=15 V [Figure: battery E on the left side. Top-left wire carries R₁ to a middle node; bottom-left wire carries R₆ to the middle bottom node. R₂ is vertic…

Question

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: R₁=15 Ω, R₂=10 Ω, R₃=20 Ω, R₄=5 Ω, R₅=25 Ω, R₆=30 Ω, E=15 V [Figure: battery E on the left side. Top-left wire carries R₁ to a middle node; bottom-left wire carries R₆ to the middle bottom node. R₂ is vertical between the two middle nodes. From the top middle node, R₃ runs to the top-right corner, R₄ runs down the right side, and R₅ runs back along the bottom to the bottom middle node.]

Accepted Answer

(2) 9/32. R₃, R₄, R₅ are in series: 20+5+25=50 Ω. This is in parallel with R₂: (50×10)/(60)=(25)/3 Ω. Add R₁ and R₆: R_eq=15+30+(25)/3=(160)/3 Ω. I=(15)/(160/3)=9/(32) A.

Step-by-step solution

Why the other options are wrong