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Simple question on current electricity
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[QUOTE="gneill, post: 5370618, member: 293536"] The OP's answer is correct. With the two internal resistances for the batteries the situation appears as follows: [ATTACH=full]95491[/ATTACH] The battery potentials are opposing, so there's 5 V of potential left to drive current. The current is thus: ##I = \frac{5~V}{(25 + 15 + 2×2.5)~Ω} = \frac{5}{45}~A = \frac{1}{9}~A## The potential difference between A and B can be found by doing a "KVL walk" from B to A. Either direction around the loop would work, but choosing the left hand path through the external resistors the path is going with the current, so the potential drops over both resistors. The total drop is: ##V_{AB} = -I×R = -\frac{1}{9}~A×40~Ω = -\frac{40}{9}~V = -4.44~V## [/QUOTE]
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Simple question on current electricity
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