CHAPTER 20 ELECTRIC CIRCUITS
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1 CHAPTER 20 ELECTRIC CIRCUITS PROBLEMS. SSM REASONING Since current is defined as charge per unit time, the current used by the portable compact disc player is equal to the charge provided by the battery pack (80 C) divided by the time in which the charge is delivered (2.0 h). SOLUTION The amount of current that the player uses in operation is determined from Equation 20.: I = q t = 80 C 2.0 h.0 h 3600 s Converts hours to seconds = A 8. REASONING As discussed in Section 20., the voltage gives the energy per unit charge. Thus, we can determine the energy delivered to the toaster by multiplying the voltage V by the charge q that flows during a time t of one minute. The charge can be obtained by solving Equation 20., I = ( q)/( t), since the current I can be obtained from Ohm s law. SOLUTION Remembering that voltage is energy per unit charge, we have Energy = V q Solving Equation 20. for q gives q = I t, which can be substituted in the previous result to give Energy = V q = VI t According to Ohm s law (Equation 20.2), the current is I = V/R, which can be substituted in the energy expression to show that Energy = VI t = V V t = V 2 t R R = ( 20 V ) 2 ( 60 s) = J 4 Ω 5. REASONING AND SOLUTION The resistance of the cable is R = V I = ρl A Since A = π r 2, the radius of the cable is r = ( 8 Ω m) 0.24 m π( V) ρli πv =.72 0 ( )( 200 A ) = m
2 Chapter 20 Problems REASONING AND SOLUTION Suppose that when the initial temperature of the wire is T 0 the resistance is R 0, and when the temperature rises to T the resistance is R. The relation between temperature and resistance is given by Equation 20.5 as R = R 0 [ + α (T T 0 )], where α is the temperature coefficient of resistivity. The initial and final resistances are related to the voltage and current as R 0 = V/I 0 and R = V/I, where the voltage V across the wire is the same in both cases. Substituting these values for R 0 and R into Equation 20.5 and solving for T, we arrive at I 0 I.50 A.30 A T = T 0 + = 20 C + α = 360 C (C ) 23. REASONING AND SOLUTION According to Equation 20.6c, the power dissipated by the iron is P = V 2 R = ( 20 V)2 24 Ω = W 25. SSM REASONING According to Equation 6.0, the energy used is Energy = Pt, where P is the power and t is the time. According to Equation 20.6a, the power is P = IV, where I is the current and V is the voltage. Thus, Energy = IVt, and we apply this result first to the drier and then to the computer. SOLUTION The energy used by the drier is 60 s Energy = Pt = IVt = (6 A)(240 V)(45 min).00 min 4243 Converts minutes to seconds For the computer, we have = J Energy = J = IVt = ( 2.7 A )( 20 V)t Solving for t, we find t = J ( 2.7 A )( 20 V ) = s = s ( ).00 h = 8.9 h 3600 s 27. REASONING AND SOLUTION We know P = V 2 /R. a. V = PR = (0.25 W)(680 Ω) = 3 V b. V = PR = (2.0 W)(680 Ω) = 37 V
3 608 ELECTRIC CIRCUITS 34. REASONING The average power is given by Equation 20.5c as P = V2 rms / R. In this expression the rms voltage V rms appears. However, we seek the peak voltage V 0. The relation between the two types of voltage is given by Equation 20.3 as V rms = V 0 / 2, so we can obtain the peak voltage by using Equation 20.3 to substitute into Equation 20.5c. SOLUTION Substituting V rms from Equation 20.3 into Equation 20.5c gives ( ) 2 P = V 2 rms R = V 0 / 2 R = V R Solving for the peak voltage V 0 gives V 0 = 2 RP = 2( 4.0Ω) ( 55 W )= 2 V 42. REASONING AND SOLUTION The equivalent resistance of the circuit is Ohm's law for the circuit gives a. Ohm's law for R gives b. Ohm's law for R 2 gives R s = R + R 2 = 36.0 Ω Ω = 54.0 Ω I = V/R s = (5.0 V)/(54.0 Ω) = A V = (0.278 A)(36.0 Ω) = 0.0 V V 2 = (0.278 A)(8.0 Ω) = 5.00 V 44. REASONING AND SOLUTION a. The equivalent resistance of the circuit is R s = 9.0 Ω Ω +.0 Ω = 5.0 Ω The current through each of the resistors is from Ohm's law b. The voltage drop across the 9.0-Ω resistor is I = (24 V)/(5.0 Ω) =.6 A V = (.6 A)(9.0 Ω) = 4 V
4 Chapter 20 Problems 609 The drop across the 5.0 Ω resistor is The drop across the.0 Ω resistor is c. The power dissipated in the 9.0 Ω resistor is Similarly, for the 5.0 Ω resistor and for the.0 Ω resistor V 2 = (.6 A)(5.0 Ω) = 8.0 V V 3 = (.6 A)(.0 Ω) =.6 V P = I 2 R = (.6 A) 2 (9.0 Ω) = 23 W P 2 = (.6 A) 2 (5.0 Ω) = 3 W P 3 = (.6 A) 2 (.0 Ω) = 2.6 W 50. REASONING AND SOLUTION The rule for combining parallel resistors is = + R P R R 2 which gives = = R 2 R P R 5 Ω or R 55 Ω 2 = 446 Ω 53. 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 calculating the power consumed by each resistor from Equation 20.6b: P = I 2 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 a. The current through the 42.0-Ω resistor is b. The power consumed by the 42.0-Ω resistor is V = IR = (3.00 A)(64.0 Ω) = 92 Ω I = V R = 92 V 42.0 Ω = 4.57 A
5 60 ELECTRIC CIRCUITS while the power consumed by the 64.0-Ω resistor is P = I 2 R = (4.57 A) 2 (42.0 Ω) = 877 W P = I 2 R = (3.00 A) 2 (64.0 Ω ) = 576 W Therefore the total power consumed by the two resistors is 877 W W = 450 W. 57. SSM REASONING The equivalent resistance of the three devices in parallel is R p, and we can find the value of R p by using our knowledge of the total power consumption of the circuit; the value of R p can be found from Equation 20.6c, P = V 2 / R p. 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 = 650 W W +250 W = 3990 W. The equivalent resistance of the circuit is R p = V 2 (20 V)2 = P 3990 W = 3.6 Ω b. The total current through the circuit is I = V R p = 20 V 3.6 Ω = 33 A The total current is larger than the rating of the circuit breaker; therefore, the breaker will open. 60. REASONING To find the current, we will use Ohm s law, together with the proper equivalent resistance. The coffee maker and frying pan are in series, so their equivalent resistance is given by Equation 20.6 as R coffee + R pan. This total resistance is in parallel with the resistance of the bread maker, so the equivalent resistance of the parallel combination can be obtained from Equation 20.7 as R p = (R coffee + R pan ) + R bread. SOLUTION Using Ohm s law and the expression developed above for R p, we find I = V = V R + p R coffee + R pan R bread = ( 20 V ) 4 Ω+6 Ω + = 23 Ω 9.2 A 67. REASONING AND SOLUTION The resistors in the small network have an equivalent resistance
6 Chapter 20 Problems 6 = R p 2.0 Ω+.0 Ω Ω+.0 Ω or R p = 2.0 Ω This resistance is in series with the 4.0-Ω resistor so the equivalent resistance of the circuit is R = 6.0 Ω. Therefore, Ohm's law gives the total current in the circuit to be I = V/R = (2 V)/(6.0 Ω) = 2.0 A This current, upon entering the parallel branch, will split in the ratios of 3:9 and 6:9, with the largest current entering the smallest resistance path. The 5.0-Ω resistor then has a current of The power dissipated in this resistor is I = (3/9)(2.0 A) P = [(3/9)(2.0 A)] 2 (5.0 Ω) = 2.2 W 68. REASONING AND SOLUTION The equivalent resistance of the initial configuration is given by /R p = 3/R or R p = R/3 The parallel part of the final configuration has a resistance of so the total equivalent resistance is R p ' = R/2 R s = R/2 + R = 3R/2 Now R s = R p Ω so 3R/2 = R/ Ω which gives R = 600 Ω
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