If an emf source is traversed from the to terminals

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Unformatted text preview: i2 r2 + iR = 0 karani (syk437) – HW13 – keto – (58945) 10. E2 + i2 r2 − iR = 0 Explanation: Recall that Kirchhoff’s loop rule states that the sum of the potential differences across all the elements around a closed circuit loop is zero. If a resistor is traversed in the direction of the current, the change in potential is −IR. If an emf source is traversed from the − to + terminals, the change in potential is +E . Apply the opposite sign for traversing the elements in the opposite direction. Thus by inspection, DCFED : E2 − iR − i2 r2 = 0 005 (part 2 of 2) 10.0 points Find the current i. Symmetry is applicable here. Let E1 = E2 = E = 8 V , r1 = r2 = r = 2.7 Ω , and R = 1.4 Ω . Correct answer: 2.90909 A. Explanation: Let : E1 = E2 = E = 8 V , r1 = r2 = r = 2 . 7 Ω , R = 1. 4 Ω . and E1 = E2 and r1 = r2 . This implies that i1 = i2 . (Loops DCFED and ABFEA have identical loop equations.) Hence the junction rule yields i1 + i2 = 2i2 = i i i2 = 2 Substituting this into the loop equation DCFED, i E2 − iR − r2 = 0 2 i= 8V r2 = 2. 7 Ω R+ 1. 4 Ω + 2 2 E2 = 2.90909 A . 4...
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