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766_PartUniversity Physics Solution

766_PartUniversity Physics Solution - 26-8 Chapter 26 2 2(d...

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26-8 Chapter 26 (d) 1 20.0 W P = (given). 2 2 2 2 2 (0.50 A) (20 ) 5.00 W P I R = = Ω = . 2 2 10 10 10 (1.0 A) (10 ) 10.0 W P I R = = Ω = . The total rate at which the resistors remove electrical energy is Resist 20 W 5 W 10 W 35.0 W P = + + = . The total rate at which the battery inputs electrical energy is Battery (3.50 A)(10.0 V) 35.0 W P I ε = = = . Resist Battery P P = , which agrees with conservation of energy. E VALUATE : The three resistors are in parallel, so the voltage for each is the battery voltage, 10.0 V. The currents in the three resistors add to give the current in the battery. 26.21. I DENTIFY : Apply Kirchhoff's point rule at point a to find the current through R . Apply Kirchhoff's loop rule to loops (1) and (2) shown in Figure 26.21a to calculate R and . E Travel around each loop in the direction shown. (a) S ET U P : Figure 26.21a E XECUTE : Apply Kirchhoff's point rule to point a : 0 so 4.00 A 6.00 A 0 I I = + = I = 2.00 A (in the direction shown in the diagram). (b) Apply Kirchhoff's loop rule to loop (1): ( )( ) ( ) 6.00 A 3.00 2.00 A 28.0 V 0 R Ω − + = ( ) 18.0 V 2.00 28.0 V 0 R Ω + = 28.0 V 18.0 V 5.00 2.00 A R = = Ω (c) Apply Kirchhoff's loop rule to loop (2): ( )( ) ( )( ) 6.00 A 3.00 4.00 A 6.00 0 Ω − Ω + = E 18.0 V 24.0 V 42.0 V = + = E E VALUATE : Can check that the loop rule is satisfied for loop (3), as a check of our work: ( )( ) ( ) 28.0 V 4.00 A 6.00 2.00 A 0 R + Ω − = E ( )( ) 28.0 V 42.0 V 24.0 V 2.00 A 5.00 0 + Ω = 52.0 V 42.0 V 10.0 V = + 52.0 V 52.0 V, = so the loop rule is satisfied for this loop. (d) I DENTIFY : If the circuit is broken at point x there can be no current in the 6.00 Ω resistor. There is now only a single current path and we can apply the loop rule to this path. S ET U P : The circuit is sketched in Figure 26.21b. Figure 26.21b E XECUTE : ( ) ( ) 28.0 V 3.00 5.00 0 I I + Ω Ω = 28.0 V 3.50 A 8.00 I = = Ω E VALUATE : Breaking the circuit at x removes the 42.0 V emf from the circuit and the current through the 3.00 Ω resistor is reduced.

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Direct-Current Circuits 26-9 26.22. I DENTIFY : Apply the loop rule and junction rule. S ET U P : The circuit diagram is given in Figure 26.22. The junction rule has been used to find the magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown currents. E XECUTE : The loop rule applied to loop (1) gives: 1 20.0 V (1.00 A)(1.00 ) (1.00 A)(4.00 ) (1.00 A)(1.00 ) (1.00 A)(6.00 ) 0 + Ω + Ω + Ω − Ω = E 1 20.0 V 1.00 V 4.00 V 1.00 V 6.00 V 18.0 V = + + = E . The loop rule applied to loop (2) gives: 2 20.0 V (1.00 A)(1.00 ) (2.00 A)(1.00 ) (2.00 A)(2.00 ) (1.00 A)(6.00 ) 0 + Ω − Ω − Ω − Ω = E 2 20.0 V 1.00 V 2.00 V 4.00 V 6.00 V 7.0 V = = E . Going from b to a along the lower branch, (2.00 A)(2.00 ) 7.0 V (2.00 A)(1.00 ) b a V V + Ω + + Ω = . 13.0 V b a V V = − ; point b is at 13.0 V lower potential than point a . E VALUATE : We can also calculate b a V V by going from b to a along the upper branch of the circuit. (1.00 A)(6.00 ) 20.0 V (1.00 A)(1.00 ) b a V V Ω + Ω = and 13.0 V b a V V = − . This agrees with b a V V calculated along a different path between b and a . Figure 26.22 26.23. I DENTIFY : Apply the junction rule at points a, b, c and d to calculate the unknown currents. Then apply the loop rule to three loops to calculate 1 2 , and . R E E (a) S ET U P : The circuit is sketched in Figure 26.23.
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766_PartUniversity Physics Solution - 26-8 Chapter 26 2 2(d...

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