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Unformatted text preview: i2 r2 + iR = 0 karani (syk437) – HW13 – keto – (58945)
10. E2 + i2 r2 − iR = 0
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.
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
i1 + i2 = 2i2 = i
Substituting this into the loop equation
E2 − iR − r2 = 0
2. 7 Ω
1. 4 Ω +
E2 = 2.90909 A . 4...
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- Spring '08