Physics Solution Manual for 1100 and 2101

0 resistor the currents that flow through the 800 and

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Unformatted text preview: med by the two resistors is 877 W + 576 W = 1450 W . ______________________________________________________________________________ 56. REASONING a. The two identical resistors, each with a resistance R, are connected in parallel across the battery, so the potential difference across each resistor is V, the potential difference provided V2 by the battery. The power P supplied to each resistor, then, is P = (Equation 20.6c). R The total power supplied by the battery is twice this amount. b. According to Equation 20.6c, the resistor which is heated until its resistance is 2R consumes only half as much power as it did initially. Because the resistance of the other resistor does not change, it consumes the same power as before. The battery supplies less power, therefore, than it did initially. We will use Equation 20.6c to determine the final power supplied. 1088 ELECTRIC CIRCUITS SOLUTION a. Since the power supplied to each resistor is P = twice as large: Ptot = 2 P = V2 (Equation 20.6c), the total power is R 2V 2 . Solving for R, we obtain R 2V 2 2 ( 25 V ) = = 130 Ω Ptot 9.6 W 2 R= b. The resistances of the resistors are now R1 = R and R2 = 2R. From Equation 20.6c, the total power output Ptot = P1 + P2 of the resistors is V 2 V 2 V 2 V 2 3V 2 3 ( 25 V ) + = + = = = 7.2 W R1 R2 R 2 R 2 R 2 (130 Ω ) ______________________________________________________________________________ 2 P = P + P2 = 1 2 57. REASONING The total power is given by Equation 20.15c as P = Vrms / R p , where Rp is the equivalent parallel resistance of the heater and the lamp. Since the total power and the rms voltage are known, we can use this expression to obtain the equivalent parallel resistance. This equivalent resistance is related to the individual resistances of the heater and –1 –1 –1 the lamp via Equation 20.17, which is R p = R heater + Rlamp . Since Rheater is given, Rlamp can be found once Rp is known. SOLUTION According to Equation 20.15c, the equivalent parallel resistance is Rp = 2 Vrms P Using this result in Equation 20.17 gives 1 1 1 1 =2 = + R p Vrms / P Rheater R lamp Rearranging this expression shows that 1 Rlamp = P 2 Vrms – 1 Rheater = 111 W (120 V ) 2 – 1 4.0 × 10 Ω 2 Therefore, Rlamp = 1 = 190 Ω 5.2 × 10 –3 Ω –1 = 5.2 × 10 –3 Ω –1 Chapter 20 Problems 58. 1089 REASONING 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 . When two resistors are connected in series, the equivalent resistance RS of the combination is greater than the resistance of either resistor alone. This can be seen directly from RS = R1 + R2 (Equation 20.16). When two resistors are connected in parallel, the equivalent resistance RP of the combination is smaller than the resistance of either resistor alone. This can be seen directly − − − by substituting values in RP 1 = R1 1 + R2 1 (Equation 20.17) or by reviewing the discussion in Section 20.7 concerning the water flow analogy for electric current in a circuit. Since the total power delivered by the battery is P = V 2 / Req and since the power and the battery voltage are the same in both cases, it follows that the equivalent resistances are also the same. But the parallel combination has an equivalent resistance RP that is smaller than RB, whereas the series combination has an equivalent resistance RS that is greater than RA. This means that RB must be greater than RA, as Diagram 1 at the right shows. If RA were greater than RB, as in Diagram 2, the equivalent resistances RS and RP would not be equal. Resistance Resistance RS RB RS = RP RB RA Diagram 1 RA RP Diagram 2 SOLUTION As discussed in the REASONING, the equivalent resistances in circuits A and B are equal. According to Equations 20.16 and 20.17, the series and parallel equivalent resistances are RS = RA + RA = 2 RA 1 1 1 = + RP RB RB or Setting the equivalent resistances equal gives RP = 1 RB 2 1090 ELECTRIC CIRCUITS 2 RA = 1 RB 2 RB =4 RA or As expected, RB is greater than RA. 59. SSM REASONING The equivalent resistance of the three devices in parallel is Rp , and we can find the value of Rp by using our knowledge of the total power consumption of the circuit; the value of Rp can be found from Equation 20.6c, P = V 2 / Rp . Ohm's law (Equation 20.2, V = IR ) can then be used to find the current through the circuit. SOLUTION a. The total power used by the circuit is P = 1650 W + 1090 W +1250 W = 3990 W. The equivalent resistance of the circuit is V 2 (120 V)2 Rp = = = 3.6 Ω P 3990 W b. The total current through the circuit is I= V 120 V = = 33 A R p 3.6 Ω This current is larger than the rating of the circuit breaker; therefore, the breaker will open . ______________________________________________________________________________ 60. REASONING AND SOLUTION The aluminum and copper portions may be viewed a being connected in parallel since the same voltage appears across them. Using a...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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