781_PartUniversity Physics Solution

781_PartUniversity Physics Solution - Direct-Current...

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Direct-Current Circuits 26-23 EXECUTE: loop (1): () ( )( ) 12 1 20.0 V 2.00 14.0 V 4.00 0 II I +− Ω −+ Ω = 6.00 4.00 6.00 A −= 3.00 2.00 3.00 A e q . ( 1 ) loop (2): ( ) ( ) 22 1 36.0 V 5.00 4.00 0 I Ω Ω = 4.00 9.00 36.0 A = e q . ( 2 ) Solving eq. (1) for 2 11 2 3 gives 1.00 A I =+ Using this in eq.(2) gives 2 3 4.00 1.00 A 9.00 36.0 A + = 8 3 9.00 40.0 A and 6.32 A. = = Then 33 1.00 A 1.00 A 6.32 A 5.21 A. = 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: 21 6.32 A 5.21 A 1.11 A. = EVALUATE: Use loop (3) to check. ( ) ( ) 20.0 V 2.00 14.0 V 36.0 V 5.00 0 + −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. IDENTIFY: Apply the loop and junction rules. SET UP: Use the currents as defined on the circuit diagram in Figure 26.64 and obtain three equations to solve for the currents. EXECUTE: 2 Left loop: 14 2( ) 0 I −− − = and 321 4 = . 121 Top loop: 2( ) 0 II I I −− + + = and 23 0 I ++ = . 12 2 Bottom loop : ( ) 2( ) 0 II I I I I −−+ + − −= and 340 . I +− = Solving these equations for the currents we find: 13 battery 1 R 2 10.0A; 6.0A; 2.0A. R I I == = = = = So the other currents are: 24 5 2 1 2 4.0 A; 4.0 A; 6.0 A. RR R I I I I =− = = − = =− + = (b) eq 14.0 V 1.40 . 10.0 A V R I = Ω EVALUATE: 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 VI R = . Figure 26.64 26.65. (a) IDENTIFY: 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 SET UP: The circuit is sketched in Figure 26.65a. Figure 26.65a EXECUTE: 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 II + 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 ΩΩ 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 VI V −+−Ω + Ω + Ω = thus ( ) 2.0 V 0.4444 A 4.00 0.22 V ab VV −= Ω = + (point a is at higher potential) EVALUATE: As a check on this calculation we also compute ab V by traveling from b to a on path (3). 10.0 V 8.0 V 2.00 1.00 2.00 b a V −+ + Ω + Ω + Ω = ( ) 2.00 V 0.4444 A 5.00 0.22 V, ab V =− + Ω =+ which checks. (b) IDENTIFY and SET UP: With points a and b connected by a wire there are three current branches, as shown in Figure 26.65b.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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781_PartUniversity Physics Solution - Direct-Current...

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