786_PartUniversity Physics Solution

786_PartUniversity Physics Solution - 26-28 Chapter 26...

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26-28 Chapter 26 EXECUTE: 1 (6.00 3.00 ) 36.0 V 0 I −Ω + Ω += 1 36.0 V 4.00 A 6.00 3.00 I == Ω+ Ω 2 (3.00 6.00 ) 36.0 V 0 I + Ω 2 36.0 V 4.00 A 3.00 6.00 I Ω To calculate ab a b VVV =− start at point b and travel to point a , adding up all the potential rises and drops along the way. We can do this by going from b up through the 3.00 Ω resistor: 21 (3.00 ) (6.00 ) ba VI I V = (4.00 A)(3.00 ) (4.00 A)(6.00 ) 12.0 V 24.0 V 12.0 V ab VV −= Ω Ω = = 12.0 V ab V (point a is 12.0 V lower in potential than point b ) EVALUATE: Alternatively, we can go from point b down through the 6.00 Ω resistor. (6.00 ) (3.00 ) I V = (4.00 A)(6.00 ) (4.00 A)(3.00 ) 24.0 V 12.0 V 12.0 V, Ω + Ω = + = which checks. (b) IDENTIFY: Now there are multiple current paths, as shown in Figure 26.73b. Use junction rule to write the current in each branch in terms of three unknown currents I 1 , I 2 , and I 3 . Apply the loop rule to three loops to get three equations for the three unknowns. The target variable is I 3 , the current through the switch. R eq is calculated from eq , R = where I is the total current that passes through the network. SET UP: The three unknown currents 12 3 , , and II I are labeled on Figure 26.73b. Figure 26.73b EXECUTE: Apply the loop rule to loops (1), (2), and (3). loop (1): 132 (6.00 ) (3.00 ) (3.00 ) 0 III + Ω + Ω = 3 2 I e q . ( 1 ) loop (2): 13 23 3 ( )(3.00 ) ( )(6.00 ) (3.00 ) 0 I −+ Ω + − Ω Ω = 1 2 3 1 61 230 s o 24 0 I I I I −− = = Use eq(1) to replace 2 : I 133 1 424 0 I −−− = 3 6 and 2 e q . ( 2 ) loop (3) (This loop is completed through the battery [not shown], in the direction from the to the terminal.): 11 3 (6.00 ) ( )(3.00 ) 36.0 V 0 I + Ω 1 3 9 3 36.0 A and 3 12.0 A I I + = e q . ( 3 ) Use eq.(2) in eq.(3) to replace 1 : I 33 3(2 ) 12.0 A 3 12.0 A/7 1.71 A I . 4 2 A 3 2 2(3.42 A) 1.71 A 5.13 A I = = The current through the switch is 3 1.71 A. I =
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Direct-Current Circuits 26-29 (c) From the results in part (a) the current through the battery is 12 3.42 A 5.13 A 8.55 A. II I = += + = The equivalent circuit is a single resistor that produces the same current through the 36.0 V battery, as shown in Figure 26.73c. 36.0 V 0 IR 36.0 V 36.0 V 4.21 8.55 A R I = == Ω Figure 26.73c EVALUATE: With the switch open (part a), point b is at higher potential than point a , so when the switch is closed the current flows in the direction from b to a . With the switch closed the circuit cannot be simplified using series and parallel combinations but there is still an equivalent resistance that represents the network. 26.74. IDENTIFY: With S open and after equilibrium has been reached, no current flows and the voltage across each capacitor is 18.0 V. When S is closed, current I flows through the 6.00 Ω and 3.00 Ω resistors. SET UP: With the switch closed, a and b are at the same potential and the voltage across the 6.00 Ω resistor equals the voltage across the 6.00 F μ capacitor and the voltage is the same across the 3.00 F capacitor and 3.00 Ω resistor. EXECUTE: (a) With an open switch: 18.0 V ab V = = E . (b) Point a is at a higher potential since it is directly connected to the positive terminal of the battery.
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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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786_PartUniversity Physics Solution - 26-28 Chapter 26...

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