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Unformatted text preview: 119 DirectCurrent Circuits 18.28 (a) Since there is not a continuous path in the
center branch, no current exists in that part
of the circuit. Then, applying Kirchhoff’s loop
rule to the outer perimeter gives +18 V + 36 V − ( 1.0 + 4.0 + 3.0 + 8.0 + 2 .0 ) Ω I = 0 or I = 18 V
+ 2.0 W 8.0 W
I zero current
5.0 W
+
ab
6.0 W
12 V
1.0 W 4.0 W
+ I I 54 V
= 3.0 A
18 Ω 3.0 W
I 36 V Now, start at point b and go around the lower loop to point a, recording changes in
potential to obtain Va − Vb = −36 V + ( 4.0 Ω + 1.0 Ω ) ( 3.0 A ) + ( 6.0 Ω + 5.0 Ω ) ( 0 ) + 12 V = −9.0 V
or ∆V ab = 9.0 V with point b at a higher potential than a
18 V
+ (b) Assume currents as shown in the modified
circuit. Applying Kirchhoff’s loop rule to the
upper loop gives I1
I − ( 11) I + 12 V − ( 7.0 ) I − ( 13 ) I1 + 18 V=0
or 18 I + 13 I1 = 30 A 13 W (1) I1 I + 11 W 12 V − ( 5.0 ) ( I1 − I ) + 36 V + ( 7.0 ) I − 12 V + ( 11) I =0
or 23 I − 5 I1 = −24 A b a
5.0 W For the lower loop, the loop rule yields 7.0 W (2) Solving equations (1) and (2) simultaneously gives I1 = 2.9 A , and I = − 0.42 A
Thus, the current in the 7.0Ω resistor is 0.42 A flowing from b to a . + 36 V I1 I1 I ...
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 Spring '09
 Physics, Current, Work

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