21_Chapter 18 HomeworkCH18 Direct-Current Circuits

21_Chapter 18 HomeworkCH18 Direct-Current Circuits - 119...

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Unformatted text preview: 119 Direct-Current 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 7.0-Ω resistor is 0.42 A flowing from b to a . + 36 V I1 I1 – I ...
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