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781_PartUniversity Physics Solution

781_PartUniversity Physics Solution - Direct-Current...

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Direct-Current Circuits 26-23 E XECUTE : loop (1): ( ) ( )( ) 1 2 1 20.0 V 2.00 14.0 V 4.00 0 I I I + Ω − + Ω = 1 2 6.00 4.00 6.00 A I I = 1 2 3.00 2.00 3.00 A I I = eq.(1) loop (2): ( ) ( )( ) 2 2 1 36.0 V 5.00 4.00 0 I I I + Ω − Ω = 1 2 4.00 9.00 36.0 A I I + = eq.(2) Solving eq. (1) for 2 1 1 2 3 gives 1.00 A I I I = + Using this in eq.(2) gives ( ) 2 2 2 3 4.00 1.00 A 9.00 36.0 A I I + + = ( ) 8 2 2 3 9.00 40.0 A and 6.32 A. I I + = = Then ( ) 2 2 1 2 3 3 1.00 A 1.00 A 6.32 A 5.21 A. I I = + = + = In summary then Current through the 2.00 Ω resistor: 1 5.21 A. I = Current through the 5.00 Ω resistor: 2 6.32 A. I = Current through the 4.00 Ω resistor: 2 1 6.32 A 5.21 A 1.11 A. I I = = E VALUATE : Use loop (3) to check. ( ) ( ) 1 2 20.0 V 2.00 14.0 V 36.0 V 5.00 0 I I + Ω − + Ω = ( )( ) ( )( ) 5.21 A 2.00 6.32 A 5.00 42.0 V Ω + Ω = 10.4 V 31.6 V 42.0 V, + = so the loop rule is satisfied for this loop. 26.64. I DENTIFY : Apply the loop and junction rules. S ET U P : Use the currents as defined on the circuit diagram in Figure 26.64 and obtain three equations to solve for the currents. E XECUTE : 1 1 2 Left loop: 14 2( ) 0 I I I = and 1 2 3 2 14 I I = . 1 2 1 Top loop : 2( ) 0 I I I I + + = and 1 2 2 3 0 I I I + + = . 1 2 1 2 2 Bottom loop : ( ) 2( ) 0 I I I I I I + + = and 1 2 3 4 0. I I I + = Solving these equations for the currents we find: 1 3 battery 1 R 2 10.0 A; 6.0 A; 2.0 A. R I I I I I I = = = = = = So the other currents are: 2 4 5 1 1 2 1 2 4.0 A; 4.0 A; 6.0 A. R R R I I I I I I I I I I = = = = = + = (b) eq 14.0 V 1.40 . 10.0 A V R I = = = Ω E VALUATE : It isn’t possible to simplify the resistor network using the rules for resistors in series and parallel. But the equivalent resistance is still defined by eq V IR = . Figure 26.64 26.65. (a) I DENTIFY : Break the circuit between points a and b means no current in the middle branch that contains the 3.00 Ω resistor and the 10.0 V battery. The circuit therefore has a single current path. Find the current, so that potential drops across the resistors can be calculated. Calculate ab V by traveling from a to b , keeping track of the potential changes along the path taken.
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26-24 Chapter 26 S ET U P : The circuit is sketched in Figure 26.65a. Figure 26.65a E XECUTE : Apply the loop rule to loop (1). ( ) ( ) 12.0 V 1.00 2.00 2.00 1.00 8.0 V 2.00 1.00 0 I I + Ω + Ω + Ω + Ω − Ω + Ω = 12.0 V 8.0 V 0.4444 A. 9.00 I = = Ω To find ab V start at point b and travel to a , adding up the potential rises and drops. Travel on path (2) shown on the diagram. The 1.00 and 3.00 Ω Ω resistors in the middle branch have no current through them and hence no voltage across them. Therefore, ( ) 10.0 V 12.0 V 1.00 1.00 2.00 ; b a V I V + Ω + Ω + Ω = thus ( )( ) 2.0 V 0.4444 A 4.00 0.22 V a b V V = Ω = + (point a is at higher potential) E VALUATE : As a check on this calculation we also compute ab V by traveling from b to a on path (3).
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