Physics Solution Manual for 1100 and 2101

Thus the chapter 20 problems 1097 resistance of each

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Unformatted text preview: ____________________________________________________________ 50. REASONING AND SOLUTION The rule for combining parallel resistors is 1 1 1 = + R P R1 R 2 which gives 1 1 1 1 1 = − = − or R 2 = 446 Ω R2 R P R1 115 Ω 155 Ω ______________________________________________________________________________ 51. SSM REASONING AND SOLUTION Since the circuit elements are in parallel, the equivalent resistance can be obtained directly from Equation 20.17: 1 1 1 1 1 = + = + or Rp = 5.3 Ω Rp R1 R2 16 Ω 8.0 Ω ______________________________________________________________________________ 52. REASONING a. The three resistors are in series, so the same current goes through each resistor: I1 = I 2 = I 3 . The voltage across each resistor is given by Equation 20.2 as V = IR. Because the current through each resistor is the same, the voltage across each is proportional to the resistance. Since R1 > R2 > R3 , we expect the ranking of the voltages to be V1 > V2 > V3 . Chapter 20 Problems 1085 b. The three resistors are in parallel, so the same voltage exists across each: V1 = V2 = V3 . The current through each resistor is given by Equation 20.2 as I = V/R. Because the voltage across each resistor is the same, the current through each is inversely proportional to the resistance. Since R1 > R2 > R3 , we expect the ranking of the currents to be I 3 > I 2 > I1 . SOLUTION a. The current through the three resistors is given by I = V/Rs, where Rs is the equivalent resistance of the series circuit. From Equation 20.16, the equivalent resistance is Rs = 50.0 Ω + 25.0 Ω + 10.0 Ω = 85.0 Ω. The current through each resistor is I1 = I 2 = I 3 = 24.0 V V = = 0.282 A Rs 85.0 Ω The voltage across each resistor is V1 = I R1 = ( 0.282 A ) ( 50.0 Ω ) = 14.1 V V2 = I R2 = ( 0.282 A )( 25.0 Ω ) = 7.05 V V3 = I R3 = ( 0.282 A )(10.0 Ω ) = 2.82 V b. The resistors are in parallel, so the voltage across each is the same as the voltage of the battery: V1 = V2 = V3 = 24.0 V The current through each resistor is equal to the voltage across each divided by the resistance: 24.0 V V I1 = = = 0.480 A R1 50.0 Ω I2 = 24.0 V V = = 0.960 A R2 25.0 Ω 24.0 V V = = 2.40 A R3 10.0 Ω ______________________________________________________________________________ I3 = 1086 ELECTRIC CIRCUITS 53. REASONING When the switch is open, no current goes to the resistor R2. Current exists only in R1, so it is the equivalent resistance. When the switch is closed, current is sent to both resistors. Since they are wired in parallel, we can use Equation 20.17 to find the equivalent resistance. Whether the switch is open or closed, the power P delivered to the circuit can be found from the relation P = V 2 / R (Equation 20.6c), where V is the battery voltage and R is the equivalent resistance. SOLUTION a. When the switch is open, there is current only in resistor R1. Thus, the equivalent resistance is R1 = 65.0 Ω . b. When the switch is closed, there is current in both resistors and, furthermore, they are wired in parallel. The equivalent resistance is 1 1 1 1 1 =+ = + RP R1 R2 65.0 Ω 96.0 Ω or RP = 38.8 Ω (20.17) c. When the switch is open, the power delivered to the circuit by the battery is given by P = V 2 / R1 , since the only resistance in the circuit is R1. Thus, the power is 2 V 2 ( 9.00 V ) P= = = 1.25 W R1 65.0 Ω (20.6) d. When the switch is closed, the power delivered to the circuit is P = V 2 / RP , where RP is the equivalent resistance of the two resistors wired in parallel: V 2 ( 9.00 V ) (20.6) = = 2.09 W RP 38.8 Ω ______________________________________________________________________________ 2 P= 54. REASONING AND SOLUTION The power P dissipated in a resistance R is given by 2 Equation 20.6c as P = V /R. The resistance R50 of the 50.0-W filament is R50 = V 2 (120.0 V) 2 = = 288 Ω P 50.0 W The resistance R100 of the 100.0-W filament is V 2 (120.0 V ) 2 = = 144 Ω P 100.0 W ______________________________________________________________________________ R100 = Chapter 20 Problems 55. 1087 SSM REASONING Since the resistors are connected in parallel, the voltage across each one is the same and can be calculated from Ohm's Law (Equation 20.2: V = IR ). Once the voltage across each resistor is known, Ohm's law can again be used to find the current in the second resistor. The total power consumed by the parallel combination can be found 2 calculating the power consumed by each resistor from Equation 20.6b: P = I R. Then, the total power consumed is the sum of the power consumed by each resistor. SOLUTION Using data for the second resistor, the voltage across the resistors is equal to V = IR = (3.00 A)(64.0 Ω)= 192 V a. The current through the 42.0-Ω resistor is I= V 192 V = = 4.57 A R 42.0 Ω b. The power consumed by the 42.0-Ω resistor is P = I 2 R = ( 4.57 A) 2 (42.0 Ω ) = 877 W while the power consumed by the 64.0-Ω resistor is P = I 2 R = (3.00 A) 2 (64.0 Ω ) = 576 W Therefore the total power consu...
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