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786_PartUniversity Physics Solution

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

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26-28 Chapter 26 E XECUTE : 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 + Ω + = 2 36.0 V 4.00 A 3.00 6.00 I = = Ω + Ω To calculate ab a b V V V = 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: 2 1 (3.00 ) (6.00 ) b a V I I V + Ω − Ω = (4.00 A)(3.00 ) (4.00 A)(6.00 ) 12.0 V 24.0 V 12.0 V a b V V = Ω − Ω = = − 12.0 V ab V = − (point a is 12.0 V lower in potential than point b ) E VALUATE : Alternatively, we can go from point b down through the 6.00 Ω resistor. 2 1 (6.00 ) (3.00 ) b a V I I V Ω + Ω = (4.00 A)(6.00 ) (4.00 A)(3.00 ) 24.0 V 12.0 V 12.0 V, a b V V = − Ω + Ω = − + = − which checks. (b) I DENTIFY : 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 , V IR = where I is the total current that passes through the network. S ET U P : The three unknown currents 1 2 3 , , and I I I are labeled on Figure 26.73b. Figure 26.73b E XECUTE : Apply the loop rule to loops (1), (2), and (3). loop (1): 1 3 2 (6.00 ) (3.00 ) (3.00 ) 0 I I I Ω + Ω + Ω = 2 1 3 2 I I I = eq.(1) loop (2): 1 3 2 3 3 ( )(3.00 ) ( )(6.00 ) (3.00 ) 0 I I I I I + Ω + Ω − Ω = 2 3 1 2 3 1 6 12 3 0 so 2 4 0 I I I I I I = = Use eq(1) to replace 2 : I 1 3 3 1 4 2 4 0 I I I I = 1 3 1 3 3 6 and 2 I I I I = = eq.(2) loop (3) (This loop is completed through the battery [not shown], in the direction from the to the + terminal.): 1 1 3 (6.00 ) ( )(3.00 ) 36.0 V 0 I I I Ω − + Ω + = 1 3 1 3 9 3 36.0 A and 3 12.0 A I I I I + = + = eq.(3) Use eq.(2) in eq.(3) to replace 1 : I 3 3 3(2 ) 12.0 A I I + = 3 12.0 A/7 1.71 A I = = 1 3 2 3.42 A I I = = 2 1 3 2 2(3.42 A) 1.71 A 5.13 A I I 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 1 2 3.42 A 5.13 A 8.55 A. I I 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 E VALUATE : 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. I DENTIFY : 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. S ET U P : 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 μ capacitor and 3.00 Ω resistor.
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