Unformatted text preview: ’ i2 r2 + iR = 0 karani (syk437) â€“ HW13 â€“ keto â€“ (58945)
10. E2 + i2 r2 âˆ’ iR = 0
Explanation:
Recall that Kirchhoï¬€â€™s loop rule states that
the sum of the potential diï¬€erences 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 âˆ’IR.
If an emf source is traversed from the âˆ’ 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 âˆ’ iR âˆ’ 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 â„¦ , and R = 1.4 â„¦ .
Correct answer: 2.90909 A.
Explanation:
Let : E1 = E2 = E = 8 V ,
r1 = r2 = r = 2 . 7 â„¦ ,
R = 1. 4 â„¦ . 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 âˆ’ iR âˆ’ r2 = 0
2
i= 8V
r2 =
2. 7 â„¦
R+
1. 4 â„¦ +
2
2
E2 = 2.90909 A . 4...
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This document was uploaded on 03/19/2014 for the course PHY 316L at University of Texas.
 Spring '08
 Schwitters

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