RCW4%20solutions

RCW4%20solutions - Physics 204, morning summer 2008...

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Unformatted text preview: Physics 204, morning summer 2008 Recitation ClassWork 4 solutions Chapter 20 Conceptual Questions #2,9,11 Concepts & Calculations #117,119,124 2. REASONING AND SOLUTION When an incandescent light bulb is turned on, the tungsten filament becomes white hot. Since the voltage is constant, the power delivered to the light bulb is given by Equation 20.6c: P = V 2 / R . From Equation 20.5, R = R0 [1 + α (T − T0 )] , where α is the temperature coefficient of resistivity and is a positive number. Thus, as the filament temperature increases, the resistance of the wire increases, and as the filament heats up, the power delivered to the bulb decreases. 9. REASONING AND SOLUTION The power rating of a 1000‐W heater specifies the power consumed when the heater is connected to an ac voltage of 120 V. The power consumed by the heater is given by Equation 20.6c: P1 = V 2 / R . When two of these heaters are connected in series, the equivalent resistance of the combination is R + R = 2R. The power consumed by two of the heaters connected in series is, therefore, P2 = V 2 /(2 R) = P1 /2 = (1000 W)/2 = 500 W . 11. REASONING AND SOLUTION A car has two headlights. The filament of one burns out. However, the other headlight stays on. We can immediately conclude that the bulbs are not connected in series. The figure below shows the series arrangement of two such bulbs. Bulb with burned out filament When the bulbs are connected in series, charges must flow through the filaments of both lights in order to have a complete circuit. Since the filament of the second bulb is burned out, charges will not be able to flow around the circuit, and neither headlight will stay on. On the other hand, if the bulbs are J connected in parallel, as shown at the right, the current will split at the junction J. Charges will be able to flow through the branch of the circuit that contains the good bulb, and that headlight will stay on. Notice that the order of the bulbs does not matter in either case. The results are the same. Bulb with burned out filament 117. CONCEPT QUESTIONS a. The resistance R of a piece of material is related to its length L and cross‐sectional area A by Equation 20.3, R = ρ L / A , where is the resistivity of the material. In order to rank the resistances, we need to evaluate L and A for each configuration in terms of L0, the unit of length. Resistance Rank ⎛ 2 ⎞ = ρ⎜ ⎟ ⎜L ⎟ L0 × 2 L0 ⎝ 0⎠ ⎛ 1 ⎞ L0 R=ρ = ρ⎜ ⎜ 8L ⎟ ⎟ 2 L0 × 4 L0 ⎝ 0⎠ ⎛ 1 ⎞ 2 L0 R=ρ = ρ⎜ ⎜ 2L ⎟ ⎟ L0 × 4 L0 ⎝ 0⎠ R=ρ a b c 4 L0 1 3 2 Therefore, the a has the largest resistance, followed by c, and then by b. b. Equation 20.2 states that the current I is equal to the voltage V divided by the resistance, I = V / R . Since the current is inversely proportional to the resistance, the largest current arises when the resistance is smallest, and vice versa. Thus, b has the largest current, followed by c, and then by a. SOLUTION a. The resistances can be found by using the results of the Concept Questions: ⎛ 2 ⎞ ⎛ ⎞ 2 −2 ⎟ = 1.50 × 10 Ω ⋅ m ⎜ ⎟ = 0.600 Ω −2 ⎜L ⎟ ⎝ 5.00 × 10 m ⎠ ⎝ 0⎠ ⎛ 1 ⎞ ⎛ ⎞ 1 −2 R = ρ⎜ ⎟ = 1.50 × 10 Ω ⋅ m ⎜ ⎟ = 0.0375 Ω −2 ⎜ 8L ⎟ 8 × 5.00 × 10 m ⎠ ⎝ ⎝ 0⎠ ⎛ 1 ⎞ ⎛ ⎞ 1 −2 R = ρ⎜ ⎟ = 1.50 × 10 Ω ⋅ m ⎜ ⎟ = 0.150 Ω −2 ⎜ 2L ⎟ 2 × 5.00 × 10 m ⎠ ⎝ ⎝ 0⎠ R = ρ⎜ ( ) ( ) ( ) b. The current in each case is given by Equation 20.2, where the value of the resistance is obtained from part (a): a b c 3.00 V V = = 5.00 A R 0.600 Ω 3.00 V V I= = = 80.0 A R 0.0375 Ω 3.00 V V I= = = 20.0 A R 0.150 Ω I= We see that the largest current is in b, with progressively smaller currents in c and a. 119. CONCEPT QUESTIONS 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 , the ranking of the voltages is V1 > V2 > V3 . 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 , the ranking of the currents is 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 + 25.0 + 10.0 85.0 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: I1 = 24.0 V V = = 0.480 A R1 50.0 Ω I2 = 24.0 V V = = 0.960 A R2 25.0 Ω I3 = 24.0 V V = = 2.40 A R3 10.0 Ω 124. CONCEPT QUESTIONS a. The total power P delivered by the battery is related to the equivalent resistance Req connected between the battery terminals and to the battery voltage V according to Equation 20.6c: P = V 2 / Req . b. We begin by recognizing that the combination of resistors in circuit A is also present in circuits B and C (see the shaded part of these circuits). In circuit B an additional resistor is in parallel with the combination from A. The equivalent resistance of resistances in parallel is always less than any of the individual resistances alone. Therefore, the equivalent resistance of circuit B is less than that of A. In circuit C an additional resistor is in series with the combination from A. The equivalent resistance of resistances in series is always greater than any of the individual resistances alone. Therefore, the equivalent resistance of circuit C is greater than that of A. We conclude then that the equivalent resistances are ranked C, A, B, with C the greatest and B the smallest. R R R R R R R R R + − V B + − V A R R R R R + − V C c. The total power delivered by the battery is P = V 2 / Req and is inversely proportional to the equivalent resistance. Since the battery voltage is the same in all three cases, this means that the power ranking is the reverse of the ranking deduced in Concept Question (b). In other words, from greatest to smallest, the total power delivered by the battery is B, A, C. SOLUTION The total power delivered by the battery is P = V 2 / Req . The voltage is given, but we must determine the equivalent resistance in each case. In circuit A each branch of the parallel combination consists of two resistances R in series. Thus, the resistance of each branch is Req = R + R = 2 R , according to Equation 20.16. The two parallel branches have an equivalent resistance that can be determined from Equation 20.17 as 1 1 1 = + RA 2 R 2 R RA = R or In circuit B the resistance of circuit A is in parallel with an additional resistance R. According to Equation 20.17, the equivalent resistance of this combination is 1 1 1 1 1 = + = + RB RA R R R or RB = 1 R 2 In circuit C the resistance of circuit A is in series with an additional resistance R. According to Equation 20.16, the equivalent resistance of this combination is RC = RA + R = R + R = 2 R We can now use P = V 2 / Req to determine the total power delivered by the battery in each case: [Circuit A ] [Circuit B ] V 2 ( 6.0 V ) P= 1 = 1 = 8.0 W R 2 ( 9.0 Ω ) 2 [Circuit C] V 2 ( 6.0 V ) P= = = 4.0 W 9.0 Ω R V 2 ( 6.0 V ) P= = = 2.0 W 2 R 2 ( 9.0 Ω ) 2 2 2 These results are consistent with our answer to Concept Question (c). ...
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This note was uploaded on 10/04/2009 for the course PHYS 750:204 taught by Professor Rollino during the Spring '09 term at Rutgers.

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