A rectangular circuit begins at the positive terminal of a battery labeled ΔV which is on the bottom side of the rectangle. The positive terminal is to the left of the negative terminal. The circuit extends left, up and then right to a resistor labeled R3 on the top side of the rectangle. To the right of R3 the circuit splits into two parallel horizontal branches.The top branch passes through a resistor labeled R1 then splits into two parallel horizontal sub-branches. The top sub-branch has a resistor labeled R2. The bottom sub-branch has a resistor labeled R4. The sub-branches then recombine.The bottom branch splits into two parallel horizontal sub-branches. There is a 4.0 Ω resistor on the top sub-branch and a 12 Ω resistor on the bottom sub-branch. The sub-branches then recombine and the bottom branch extends to the right to a 2.0 Ω resistor.The top and bottom branches then recombine and the circuit extends down and to the left until it reaches the negative terminal of the battery. Find the current in the 12-Ω resistor in the figure below. (Assume R1 = R3 = 1.6 Ω, R2 = R4 = 6.4 Ω, ΔV = 23 V.) ????A

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Answered: A rectangular circuit begins at the…

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter28: Direct-current Circuits

Section: Chapter Questions

Problem 28.10OQ: The terminals of a battery are connected across two resistors in parallel. The resistances of the...

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A rectangular circuit begins at the positive terminal of a battery labeled ΔV which is on the bottom side of the rectangle. The positive terminal is to the left of the negative terminal. The circuit extends left, up and then right to a resistor labeled R₃ on the top side of the rectangle. To the right of R₃ the circuit splits into two parallel horizontal branches.

The top branch passes through a resistor labeled R₁ then splits into two parallel horizontal sub-branches. The top sub-branch has a resistor labeled R₂. The bottom sub-branch has a resistor labeled R₄. The sub-branches then recombine.
The bottom branch splits into two parallel horizontal sub-branches. There is a 4.0 Ω resistor on the top sub-branch and a 12 Ω resistor on the bottom sub-branch. The sub-branches then recombine and the bottom branch extends to the right to a 2.0 Ω resistor.

The top and bottom branches then recombine and the circuit extends down and to the left until it reaches the negative terminal of the battery.

Find the current in the 12-Ω resistor in the figure below. (Assume R₁ = R₃ = 1.6 Ω, R₂ = R₄ = 6.4 Ω, ΔV = 23 V.)
????A

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