Chapter 26 HW Solutions

Chapter 26 HW Solutions - PHYS 241 Ch. 26 Homework Solutions

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PHYS 241 Ch. 26 Homework Solutions 26.7.I DENTIFY : First do as much series-parallel reduction as possible. S ET U P : The 45.0-Ω and 15.0-Ω resistors are in parallel, so first reduce them to a single equivalent resistance. Then find the equivalent series resistance of the circuit. E XECUTE : 1/ R p = 1/(45.0 Ω) + 1/(15.0 Ω) and R p = 11.25 Ω. The total equivalent resistance is 18.0 Ω + 11.25 Ω + 3.26 Ω = 32.5 Ω. Ohm’s law gives I = (25.0 V)/(32.5 Ω) = 0.769 A. E VALUATE : The circuit appears complicated until we realize that the 45.0-Ω and 15.0-Ω resistors are in parallel. 26.11. I DENTIFY : For resistors in parallel, the voltages are the same and the currents add. eq 1 2 1 1 1 R R R = + so 1 2 eq 1 2 , R R R R R = + For resistors in series, the currents are the same and the voltages add. eq 1 2 R R R = + . S ET U P : The rules for combining resistors in series and parallel lead to the sequences of equivalent circuits shown in Figure 26.11. E XECUTE : eq 5.00 R = . In Figure 26.11c, 60.0 V 12.0 A 5.00 I = = . This is the current through each of the resistors in Figure 26.11b. 12 12 (12.0 A)(2.00 ) 24.0 V V IR = = Ω = . 34 34 (12.0 A)(3.00 ) 36.0 V V IR = = Ω = . Note that 12 34 60.0 V V V + = . 12 V is the voltage across 1 R and across 2 R , so 12 1 1 24.0 V 8.00 A 3.00 V I R = = = and 12 2 2 24.0 V 4.00 A 6.00 V I R = = = . 34 V is the voltage across 3 R and across 4 R , so 34 3 3 36.0 V 3.00 A 12.0 V I R = = = and 34 4 4 36.0 V 9.00 A 4.00 V I R = = = . E VALUATE : Note that 1 2 3 4 I I I I + = + . Figure 26.11 26.13. I DENTIFY : In both circuits, with and without R 4 , replace series and parallel combinations of resistors by their equivalents. Calculate the currents and voltages in the equivalent circuit and infer from this the currents and voltages in the original circuit. Use 2 P I R = to calculate the power dissipated in each bulb. (a) S ET U P : The circuit is sketched in Figure 26.13a. E XECUTE : 2 3 4 , , and R R R are in parallel, so their equivalent resistance eq R is given by eq 2 3 4 1 1 1 1 R R R R = + + Figure 26.13a 1
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eq eq 1 3 and 1.50 . 4.50 R R = = The equivalent circuit is drawn in Figure 26.13b. ( 29 1 eq 0 I R R - + = E 1 eq I R R = + E Figure 26.13b 1 9.00 V 1.50 A and 1.50 A 4.50 1.50 I I = = = Ω + Then ( 29 ( 29 1 1 1 1.50 A 4.50 6.75 V V I R = = Ω = ( 29 ( 29 eq eq eq eq 1.50 A, 1.50 A 1.50 2.25 V I V I R = = = Ω = For resistors in parallel the voltages are equal and are the same as the voltage across the equivalent resistor, so 2 3 4 2.25 V. V V V = = = 2 3 4 2 3 4 2 3 4 2.25 V 0.500 A, 0.500 A, 0.500 A 4.50 V V V I I I R R R = = = = = = = E VALUATE : Note that 2 3 4 1.50 A, I I I + + = which is eq . I For resistors in parallel the currents add and their sum is the current through the equivalent resistor. (b) S
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Chapter 26 HW Solutions - PHYS 241 Ch. 26 Homework Solutions

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