Simple Resistive Circuits-Electrical Circuit Analysis-Solution Manual, Exercises of Electrical Circuit Analysis

Download Simple Resistive Circuits-Electrical Circuit Analysis-Solution Manual and more Exercises Electrical Circuit Analysis in PDF only on Docsity! 3 Simple Resistive Circuits Assessment Problems AP 3.1 Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains. Begin by combining the 6 Ω resistor and the 10 Ω resistor in series: 6 Ω + 10 Ω = 16 Ω Now combine this 16 Ω resistor in parallel with the 64 Ω resistor: 16 Ω‖64 Ω = (16)(64) 16 + 64 = 1024 80 = 12.8 Ω This equivalent 12.8 Ω resistor is in series with the 7.2 Ω resistor: 12.8 Ω + 7.2 Ω = 20 Ω Finally, this equivalent 20 Ω resistor is in parallel with the 30 Ω resistor: 20 Ω‖30 Ω = (20)(30) 20 + 30 = 600 50 = 12 Ω Thus, the simplified circuit is as shown: 3–1 docsity.com 3–2 CHAPTER 3. Simple Resistive Circuits [a] With the simplified circuit we can use Ohm’s law to find the voltage across both the current source and the 12 Ω equivalent resistor: v = (12 Ω)(5 A) = 60 V [b] Now that we know the value of the voltage drop across the current source, we can use the formula p = −vi to find the power associated with the source: p = −(60 V)(5 A) = −300 W Thus, the source delivers 300 W of power to the circuit. [c] We now can return to the original circuit, shown in the first figure. In this circuit, v = 60 V, as calculated in part (a). This is also the voltage drop across the 30 Ω resistor, so we can use Ohm’s law to calculate the current through this resistor: iA = 60 V 30 Ω = 2 A Now write a KCL equation at the upper left node to find the current iB: −5 A + iA + iB = 0 so iB = 5 A − iA = 5 A − 2 A = 3 A Next, write a KVL equation around the outer loop of the circuit, using Ohm’s law to express the voltage drop across the resistors in terms of the current through the resistors: −v + 7.2iB + 6iC + 10iC = 0 So 16iC = v − 7.2iB = 60 V − (7.2)(3) = 38.4 V Thus iC = 38.4 16 = 2.4 A Now that we have the current through the 10 Ω resistor we can use the formula p = Ri2 to find the power: p10 Ω = (10)(2.4)2 = 57.6 W AP 3.2 [a] We can use voltage division to calculate the voltage vo across the 75 kΩ resistor: vo(no load) = 75,000 75,000 + 25,000 (200 V) = 150 V docsity.com Problems 3–5 [c] We can find the power dissipated by the 50 Ω resistor if we can find the current in this resistor. We can use current division to find this current from the current in the 40 Ω resistor, but first we need to calculate the equivalent resistance of the 20 Ω branch and the 30 Ω branch: 20 Ω‖30 Ω = (20)(30) 20 + 30 = 12 Ω Current division gives: i50Ω = 12 12 + 50 + 10 (0.5 A) = 0.08333 A Thus, p50Ω = (50)(0.08333)2 = 0.34722 W = 347.22 mW AP 3.5 [a] We can find the current i using Ohm’s law: i = 1 V 100 Ω = 0.01 A = 10 mA [b] Rm = 50 Ω‖5.555 Ω = 5 Ω We can use the meter resistance to find the current using Ohm’s law: imeas = 1 V 100 Ω + 5 Ω = 0.009524 = 9.524 mA AP 3.6 [a] docsity.com 3–6 CHAPTER 3. Simple Resistive Circuits Use voltage division to find the voltage v: v = 75,000 75,000 + 15,000 (60 V) = 50 V [b] The meter resistance is a series combination of resistances: Rm = 149,950 + 50 = 150,000 Ω We can use voltage division to find v, but first we must calculate the equivalent resistance of the parallel combination of the 75 kΩ resistor and the voltmeter: 75,000 Ω‖150,000 Ω = (75,000)(150,000) 75,000 + 150,000 = 50 kΩ Thus, vmeas = 50,000 50,000 + 15,000 (60 V) = 46.15 V AP 3.7 [a] Using the condition for a balanced bridge, the products of the opposite resistors must be equal. Therefore, 100Rx = (1000)(150) so Rx = (1000)(150) 100 = 1500 Ω = 1.5 kΩ [b] When the bridge is balanced, there is no current flowing through the meter, so the meter acts like an open circuit. This places the following branches in parallel: The branch with the voltage source, the branch with the series combination R1 and R3 and the branch with the series combination of R2 and Rx. We can find the current in the latter two branches using Ohm’s law: iR1,R3 = 5 V 100 Ω + 150 Ω = 20 mA; iR2,Rx = 5 V 1000 + 1500 = 2 mA We can calculate the power dissipated by each resistor using the formula p = Ri2: p100Ω = (100 Ω)(0.02 A)2 = 40 mW p150Ω = (150 Ω)(0.02 A)2 = 60 mW p1000Ω = (1000 Ω)(0.002 A)2 = 4 mW p1500Ω = (1500 Ω)(0.002 A)2 = 6 mW Since none of the power dissipation values exceeds 250 mW, the bridge can be left in the balanced state without exceeding the power-dissipating capacity of the resistors. docsity.com Problems 3–7 AP 3.8 Convert the three Y-connected resistors, 20 Ω, 10 Ω, and 5 Ω to three ∆-connected resistors Ra, Rb, and Rc. To assist you the figure below has both the Y-connected resistors and the ∆-connected resistors Ra = (5)(10) + (5)(20) + (10)(20) 20 = 17.5 Ω Rb = (5)(10) + (5)(20) + (10)(20) 10 = 35 Ω Rc = (5)(10) + (5)(20) + (10)(20) 5 = 70 Ω The circuit with these new ∆-connected resistors is shown below: From this circuit we see that the 70 Ω resistor is parallel to the 28 Ω resistor: 70 Ω‖28 Ω = (70)(28) 70 + 28 = 20 Ω Also, the 17.5 Ω resistor is parallel to the 105 Ω resistor: 17.5 Ω‖105 Ω = (17.5)(105) 17.5 + 105 = 15 Ω Once the parallel combinations are made, we can see that the equivalent 20 Ω resistor is in series with the equivalent 15 Ω resistor, giving an equivalent resistance docsity.com 3–10 CHAPTER 3. Simple Resistive Circuits The simplified circuit is shown below: [b] The 9 kΩ, 18 kΩ, and 6 kΩ resistors are in parallel. The simplified circuit is shown below: [c] The 600 Ω, 200 Ω, and 300 Ω resistors are in series. The simplified circuit is shown below: P 3.3 [a] p4Ω = i2s4 = (12) 24 = 576 W p18Ω = (4)218 = 288 W p3Ω = (8)23 = 192 W p6Ω = (8)26 = 384 W [b] p120V(delivered) = 120is = 120(12) = 1440 W [c] pdiss = 576 + 288 + 192 + 384 = 1440 W P 3.4 [a] From Ex. 3-1: i1 = 4 A, i2 = 8 A, is = 12 A at node x: −12 + 4 + 8 = 0, at node y: 12 − 4 − 8 = 0 docsity.com Problems 3–11 [b] v1 = 4is = 48 V v3 = 3i2 = 24 V v2 = 18i1 = 72 V v4 = 6i2 = 48 V loop abda: −120 + 48 + 72 = 0, loop bcdb: −72 + 24 + 48 = 0, loop abcda: −120 + 48 + 24 + 48 = 0 P 3.5 Always work from the side of the circuit furthest from the source. Remember that the current in all series-connected circuits is the same, and that the voltage drop across all parallel-connected resistors is the same. [a] Req = 6 + 12 + [4‖(9 + 7)] = 18 + (4‖16) = 18 + 3.2 = 21.2 Ω [b] Req = 4 k + [10 k‖(3 k + 5 k + 7 k)] = 4 k + (10 k‖15 k) = 4 k + 6 k = 10 kΩ [c] Req = (300 + 400 + 500) + (600‖1200) = 1200 + 400 = 1600 Ω P 3.6 Always work from the side of the circuit furthest from the source. Remember that the current in all series-connected circuits is the same, and that the voltage drop across all parallel-connected resistors is the same. [a] Req = 18 + (100‖25‖(22 + (10‖40))) = 18 + (20‖(22 + 8) = 18 + 12 = 30 Ω [b] Req = 10 k‖(5 k + 2 k + (9 k‖18 k‖6 k)) = 10 k‖(7 k + 3 k) = 10 k‖10 k = 5 kΩ [c] Req = 600‖200‖300‖(250 + 150) = 600‖200‖300‖400 = 80 Ω P 3.7 [a] Req = 12 + (24‖(30 + 18)) + 10 = 12 + (24‖48) + 10 = 12 + 16 + 10 = 38 Ω [b] Req = 4 k‖30 k‖60 k‖(1.2 k + (7.2 k‖2.4 k) + 2 k) = 4 k‖30 k‖60 k‖(3.2 k + 1.8 k) = 4 k‖30 k‖60 k‖5 k = 2 kΩ P 3.8 [a] 5‖20 = 100/25 = 4 Ω 5‖20 + 9‖18 + 10 = 20 Ω 9‖18 = 162/27 = 6 Ω 20‖30 = 600/50 = 12 Ω Rab = 5 + 12 + 3 = 20 Ω docsity.com 3–12 CHAPTER 3. Simple Resistive Circuits [b] 5 + 15 = 20 Ω 30‖20 = 600/50 = 12 Ω 20‖60 = 1200/80 = 15 Ω 3‖6 = 18/9 = 2 Ω 15 + 10 = 25 Ω 3‖6 + 30‖20 = 2 + 12 = 14 Ω 25‖75 = 1875/100 = 18.75 Ω 26‖14 = 364/40 = 9.1 Ω 18.75 + 11.25 = 30 Ω Rab = 2.5 + 9.1 + 3.4 = 15 Ω [c] 3 + 5 = 8 Ω 60‖40 = 2400/100 = 24 Ω 8‖12 = 96/20 = 4.8 Ω 24 + 6 = 30 Ω 4.8 + 5.2 = 10 Ω 30‖10 = 300/40 = 7.5 Ω 45 + 15 = 60 Ω Rab = 1.5 + 7.5 + 1.0 = 10 Ω P 3.9 [a] For circuit (a) Rab = 360‖(90 + 120‖(160 + 200)) = 360‖(90 + (120‖360)) = 360‖(90 + 90) = 360‖180 = 120 Ω For circuit (b) 1 Re = 1 20 + 1 15 + 1 20 + 1 4 + 1 12 = 30 60 = 1 2 Re = 2 Ω Re + 16 = 18 Ω 18‖18 = 9 Ω Rab = 10 + 8 + 9 = 27 Ω For circuit (c) 15‖30 = 10 Ω 10 + 20 = 30 Ω 60‖30 = 20 Ω 20 + 10 = 30 Ω 30‖80‖(40 + 20) = 30‖80‖60 = 16 Ω Rab = 16 + 24 + 10 = 50 Ω docsity.com Problems 3–15 Now use the voltage specification: R2 R2 + 3600 (75) = 15 Thus, R2 = 900 Ω R1 = 1600 Ω and R2 = 400 Ω are the smallest values of resistors that satisfy the 1 W specification. P 3.14 Use voltage division to determine R2 from the no-load voltage specification: 6 V = R2 (R2 + 40) (18 V); so 18R2 = 6(R2 + 40) Thus, 12R2 = 240 so R2 = 240 12 = 20 Ω Now use voltage division again, this time to determine the value of Re, the parallel combination of R2 and RL. We use the loaded voltage specification: 4 V = Re (40 + Re) (18 V) so 18Re = 4(40 + Re) Thus, 14Re = 160 so Re = 160 14 = 11.43 Ω Now use the definition Re to calculate the value of RL given that R2 = 20 Ω: Re = 20RL 20 + RL = 11.43 so 20RL = 11.43(RL + 20) Therefore, 8.57RL = 228.6 and RL = 226.8 8.57 = 26.67 Ω P 3.15 [a] From the constraint on the no-load voltage, R2 R1 + R2 (40) = 8 so R1 = 4R2 From the constraint on the loaded voltage divider: 7.5 = 3600R2 3600 + R2 R1 + 3600R2 3600 + R2 (40) = 3600R2 3600 + R2 4R2 + 3600R2 3600 + R2 (40) docsity.com 3–16 CHAPTER 3. Simple Resistive Circuits = 3600R2 4R2(3600 + R2) + 3600R2 (40) = 144,000R2 4R22 + 18,000R2 So, 144,000 4R2 + 18,000 = 7.5 ·. . R2 = 300 Ω and R1 = 4R2 = 1200 Ω [b] Power dissipated in R1 will be maximum when the voltage across R1 is maximum. This will occur under load conditions. vR1 = 40 − 7.5 = 32.5 V; PR1 = (32.5)2 1200 = 880.2 mW So specify a 1 W power rating for the resistor R1. The power dissipated in R2 will be maximum when the voltage drop across R2 is maximum. This occurs under no-load conditions with vo = 8 V. PR2 = (8)2 300 = 213.3 m W So specify a 1/4 W power rating for the resistor R2. P 3.16 Refer to the solution of Problem 3.15. The divider will reach its dissipation limit when the power dissipated in R1 equals 1 W So (v2R1/1200) = 1; vR1 = 34.641 V vo = 40 − 34.641 = 5.359 V Therefore, Re 1200 + Re (40) = 5.359, and Re = 185.641 Ω 1200RL 1200 + RL = 185.641 ·. . RL = 219.62 Ω P 3.17 [a] 120 kΩ + 30 kΩ = 150 kΩ 75 kΩ‖150 kΩ = 50 kΩ vo1 = 240 (25,000 + 50,000) (50,000) = 160 V vo = 120,000 (150,000) (vo1) = 128 V, vo = 128 V docsity.com Problems 3–17 [b] i = 240 100,000 = 2.4 mA 75,000i = 180 V vo = 120,000 150,000 (180) = 144 V; vo = 144 V [c] It removes loading effect of second voltage divider on the first voltage divider. Observe that the open circuit voltage of the first divider is v′o1 = 75,000 (100,000) (240) = 180 V Now note this is the input voltage to the second voltage divider when the current controlled voltage source is used. P 3.18 (24)2 R1 + R2 + R3 = 36, Therefore, R1 + R2 + R3 = 16 Ω (R1 + R2)24 (R1 + R2 + R3) = 12 Therefore, 2(R1 + R2) = R1 + R2 + R3 Thus, R1 + R2 = R3; 2R3 = 16; R3 = 8 Ω R2(24) R1 + R2 + R3 = 6 4R2 = R1 + R2 + R3 so R2 = R3/2 = 4 Ω R2 = 4 Ω; R1 = 16 − 8 − 4 = 4 Ω P 3.19 Note – in the problem description, the first equation defines R1 not RL. [a] At no load: vo = kvs = R2 R1 + R2 vs. At full load: vo = αvs = Re R1 + Re vs, where Re = RoR2 Ro + R2 Therefore k = R2 R1 + R2 and R1 = (1 − k) k R2 α = Re R1 + Re and R1 = (1 − α) α Re docsity.com 3–20 CHAPTER 3. Simple Resistive Circuits P 3.22 [a] Using voltage division, v18Ω = 18 18 + 30 (40) = 15 V positive at the top [b] Using current division, i30Ω = 24 24 + 30 + 18 (60 × 10−3) = 20 mA flowing from right to left [c] The 9 mA current in the 1.2 kΩ resistor is also the current in the 2 kΩ resistor. It then divides among the 4 kΩ, 30 kΩ, and 60 kΩ resistors. 4 kΩ‖60 kΩ = 3.75 kΩ Using current division, i30 kΩ = 3.75 k 30 k + 3.75 k (9 × 10−3) = 1 m A, flowing bottom to top [d] docsity.com Problems 3–21 The voltage drop across the 4 kΩ resistor is the same as the voltage drop across the series combination of the 1.2 kΩ, the (7.2 k‖2.4 k)Ω combined resistor, and the 2 kΩ resistor. Note that 7.2 k‖2.4 k = (7200)(2400) 9600 = 1.8 kΩ Using voltage division, vo = 1800 1200 + 1800 + 2000 (50) = 18 V positive at the top P 3.23 [a] First, note the following: 18‖9 = 6 Ω; 20‖5 = 4 Ω; and the voltage drop across the 18 Ω resistor is the same as the voltage drop across the parallel combination of the 18 Ω and 9 Ω resistors. Using voltage division, vo = 6 6 + 4 + 10 (0.1 V) = 30 mV positive at the left [b] The equivalent resistance of the 5 Ω, 15 Ω, and 60 Ω resistors is Re = (5 + 15)‖60 = 15 Ω Using voltage division to find the voltage across the equivalent resistance, vRe = 15 15 + 10 (10) = 6 V Using voltage division again, vo = 15 5 + 15 (6) = 4.5 V positive at the top docsity.com 3–22 CHAPTER 3. Simple Resistive Circuits [c] Find equivalent resistance on the right side Rr = 5.2 + (12)(5 + 3) (12 + 3 + 5) = 10 Ω Find voltage bottom to top across Rr (10)(3) = 30 V Find the equivalent resistance on the left side Rl = 6 + (40)(45 + 15) (40 + 45 + 15) = 30 Ω The current in the 6 Ω is i6 Ω = 30 30 = 1 A left to right Use current division to find io io = (1) ( 40 40 + 15 + 45 ) = 0.4 A bottom to top P 3.24 [a] v20k = 20 20 + 5 (45) = 36 V v90k = 90 90 + 60 (45) = 27 V vx = v20k − v90k = 36 − 27 = 9 V [b] v20k = 20 25 (Vs) = 0.8Vs v90k = 90 150 (Vs) = 0.6Vs vx = 0.8Vs − 0.6Vs = 0.2Vs P 3.25 150‖75 = 50 Ω The equivalent resistance to the right of the 90 Ω resistor is (50 + 40)‖(60 + 30) = 45 Ω docsity.com Problems 3–25 i1 = (3)(40) (60) = 2 A; vx = 8i1 = 16 V vg = 20i1 = 40 V v60 = vg − vx = 24 V Pdevice = 242 60 + 162 10 + 402 40 = 75.2 W P 3.31 [a] The model of the ammeter is an ideal ammeter in parallel with a resistor whose resistance is given by Rs = 100 µV 10 µA = 10 Ω. We can calculate the current through the real meter using current division: im = (10/99) 10 + (10/99) (imeas) = 10 990 + 10 (imeas) = 1 100 imeas [b] Rs = 100 µV 10 µA = 10 Ω. im = (100/999,990) 10 + (100/999,990) (imeas) = 1 100,000 (imeas) [c] Yes P 3.32 Measured value: 60‖20.1 = 15.056 Ω ig = 50 (15.056 + 10) = 1.9955 A; imeas = (1.9955) 60 80.1 = 1.495 A True value: 60‖20 = 15 Ω ig = 50 (15 + 10) = 50 25 = 2.0 A; itrue = (2) (60 80 ) = 1.5 A % error = [1.495 1.5 − 1 ] × 100 = −0.3488% docsity.com 3–26 CHAPTER 3. Simple Resistive Circuits P 3.33 Begin by using current division to find the actual value of the current io: itrue = 15 15 + 45 (50 mA) = 12.5 mA imeas = 15 15 + 45 + 0.1 (50 mA) = 12.48 mA % error = [12.48 12.5 − 1 ] 100 = −0.1664% P 3.34 For all full-scale readings the total resistance is RV + Rmovement = full-scale reading 10−3 We can calculate the resistance of the movement as follows: Rmovement = 20 mV 1 mA = 20 Ω Therefore, RV = 1000 (full-scale reading) − 20 [a] RV = 1000(50) − 20 = 49, 980 Ω [b] RV = 1000(5) − 20 = 4980 Ω [c] RV = 1000(0.25) − 20 = 230 Ω [d] RV = 1000(0.025) − 20 = 5 Ω P 3.35 [a] vmeas = (50 × 10−3)[15‖45‖(4980 + 20)] = 0.5612 V [b] vtrue = (50 × 10−3)(15‖45) = 0.5625 V % error = (0.5612 0.5625 − 1 ) × 100 = −0.23% P 3.36 Original meter: Re = 50 × 10−3 5 = 0.01 Ω Modified meter: Re = (0.02)(0.01) 0.03 = 0.00667 Ω ·. . (Ifs)(0.00667) = 50 × 10−3 ·. . Ifs = 7.5 A docsity.com Problems 3–27 P 3.37 At full scale the voltage across the shunt resistor will be 100 mV; therefore the power dissipated will be PA = (100 × 10−3)2 RA Therefore RA ≥ (100 × 10 −3)2 0.25 = 40 mΩ Otherwise the power dissipated in RA will exceed its power rating of 0.25 W When RA = 40 mΩ, the shunt current will be iA = 100 × 10−3 40 × 10−3 = 2.5 A The measured current will be imeas = 2.5 + 0.001 = 2.501 A·. . Full-scale reading is for practical purposes is 2.5 A P 3.38 The current in the shunt resistor at full-scale deflection is iA = ifullscale − 20 × 10−6 The voltage across RA at full-scale deflection is always 10 mV, therefore RA = 10 × 10−3 ifullscale − 2 × 10−3 = 10 1000ifs − 0.02 [a] RA = 10 10,000 − 0.02 = 1 mΩ [b] RA = 10 1000 − 0.02 = 10 mΩ [c] RA = 10 100 − 0.02 = 1 Ω [d] RA = 10 0.1 − 0.02 = 125 Ω P 3.39 [a] docsity.com 3–30 CHAPTER 3. Simple Resistive Circuits P 3.44 Note – the upper terminal of the voltmeter should be labeled 820 V, not 300 V. [a] Rmeter = 360 kΩ + 200 kΩ‖50 kΩ = 400 kΩ 400‖600 = 240 kΩ Vmeter = 240 300 (300) = 240 V [b] What is the percent error in the measured voltage? True value = 600 660 (300) = 272.73 V % error = ( 240 272.73 − 1 ) 100 = −12% P 3.45 [a] R1 = 100 V 2 mA = 50 kΩ R2 = 10 V 2 mA = 5 kΩ R3 = 1 V 2 mA = 500 Ω [b] Let ia = actual current in the movement id = design current in the movement Then % error = ( ia id − 1 ) 100 For the 100 V scale: ia = 100 50,000 + 25 = 100 50,025 , id = 100 50,000 ia id = 50,000 50,025 = 0.9995 % error = (0.9995 − 1)100 = −0.05% For the 10 V scale: ia id = 5000 5025 = 0.995 % error = (0.995 − 1.0)100 = −0.5% For the 1 V scale: ia id = 500 525 = 0.9524 % error = (0.9524 − 1.0)100 = −4.76% P 3.46 [a] Rmovement = 50 Ω R1 + Rmovement = 30 1 × 10−3 = 30 kΩ ·. . R1 = 29,950 Ω R2 + R1 + Rmovement = 150 1 × 10−3 = 150 kΩ ·. . R2 = 120 kΩ docsity.com Problems 3–31 R3 + R2 + R1 + Rmovement = 300 1 × 10−3 = 300 kΩ ·. . R3 = 150 kΩ docsity.com 3–32 CHAPTER 3. Simple Resistive Circuits [b] imove = 288 300 (1) = 0.96 mA v1 = (0.96 m)(150 k) = 144 V i1 = 144 750 k = 0.192 mA i2 = imove + i1 = 0.96 m + 0.192 m = 1.152 mA vmeas = vx = 144 + 150i2 = 316.8 V [c] v1 = 150 V; i2 = 1 m + 0.20 m = 1.20 mA i1 = 150/750,000 = 0.20 mA ·. . vmeas = vx = 150 + (150 k)(1.20 m) = 330 V P 3.47 From the problem statement we have 50 = Vs(10) 10 + Rs (1) Vs in mV; Rs in MΩ 48.75 = Vs(6) 6 + Rs (2) [a] From Eq (1) 10 + Rs = 0.2Vs ·. . Rs = 0.2Vs − 10 Substituting into Eq (2) yields 48.75 = 6Vs 0.2Vs − 6 or Vs = 52 mV [b] From Eq (1) 50 = 520 10 + Rs or 50Rs = 20 So Rs = 400 kΩ docsity.com Problems 3–35 i1 = 30 V 6000 Ω = 5 mA i2 = 45 V 12,000 Ω = 3.75 mA id = i1 − i2 = 5 mA − 3.75 mA = 1.25 mA P 3.51 Note the bridge structure is balanced, that is 15 × 5 = 25 × 3, hence there is no current in the 5 kΩ resistor. It follows that the equivalent resistance of the circuit is Req = 0.750 + 11.25 = 12 kΩ The source current is 192/12,000 = 16 mA. The current down through the 3 kΩ resistor is i3k = 16 30 48 = 10 mA ·. . p3k = (10 × 10−3)2(3 × 103) = 300 mW P 3.52 In order that all four decades (1, 10, 100, 1000) that are used to set R3 contribute to the balance of the bridge, the ratio R2/R1 should be set to 0.001. P 3.53 Begin by transforming the Y-connected resistors (10 Ω, 40 Ω, 50 Ω) to ∆-connected resistors. Both the Y-connected and ∆-connected resistors are shown below to assist in using Eqs. 3.44 – 3.46: Now use Eqs. 3.44 – 3.46 to calculate the values of the ∆-connected resistors: R1 = (40)(10) 10 + 40 + 50 = 4 Ω; R2 = (50)(10) 10 + 40 + 50 = 5 Ω; R3 = (40)(50) 10 + 40 + 50 = 20 Ω The transformed circuit is shown below: docsity.com 3–36 CHAPTER 3. Simple Resistive Circuits The equivalent resistance seen by the 24 V source can be calculated by making series and parallel combinations of the resistors to the right of the 24 V source: Req = (15 + 5)‖(4 + 1) + 20 = 20‖5 + 20 = 4 + 20 = 24 Ω Therefore, the current i in the 24 V source is given by i = 24 V 24 Ω = 1 A Use current division to calculate the currents i1 and i2. Note that the current i1 flows in the branch containing the 15 Ω and 5 Ω series connected resistors, while the current i2 flows in the parallel branch that contains the series connection of the 1 Ω and 4 Ω resistors: i1 = 1 + 4 1 + 4 + 15 + 5 (i) = 5 25 (1 A) = 0.2 A, and i2 = 1 A − 0.2 A = 0.8 A Now use KVL and Ohm’s law to calculate v1. Note that v1 is the sum of the voltage drop across the 4 Ω resistor, 4i2, and the voltage drop across the 20 Ω resistor, 20i: v1 = 4i2 + 20i = 4(0.8 A) + 20(1 A) = 3.2 + 20 = 23.2 V Finally, use KVL and Ohm’s law to calculate v2. Note that v2 is the sum of the voltage drop across the 5 Ω resistor, 5i1, and the voltage drop across the 20 Ω resistor, 20i: v2 = 5i1 + 20i = 5(0.2 A) + 20(1 A) = 1 + 20 = 21 V P 3.54 [a] Calculate the values of the Y-connected resistors that are equivalent to the 10 Ω, 40 Ω, and 50Ω ∆-connected resistors: RX = (10)(50) 10 + 40 + 50 = 5 Ω; RY = (40)(50) 10 + 40 + 50 = 20 Ω; RZ = (10)(40) 10 + 40 + 50 = 4 Ω Replacing the R2—R3—R4 delta with its equivalent Y gives docsity.com Problems 3–37 Now calculate the equivalent resistance Rab by making series and parallel combinations of the resistors: Rab = 13 + 5 + [(4 + 8)‖(20 + 4)] + 7 = 33 Ω [b] Calculate the values of the ∆-connected resistors that are equivalent to the 8 Ω, 10 Ω, and 40 Ω Y-connected resistors: RX = (10)(40) + (40)(8) + (8)(10) 8 = 800 8 = 100 Ω RY = (10)(40) + (40)(8) + (8)(10) 10 = 800 10 = 80 Ω RZ = (10)(40) + (40)(8) + (8)(10) 40 = 800 40 = 20 Ω Replacing the R2, R4, R5 wye with its equivalent ∆ gives Make series and parallel combinations of the resistors to find the equivalent resistance Rab: 100 Ω‖50 Ω = 33.33 Ω; 80 Ω‖4 Ω = 3.81 Ω ·. . 100‖50 + 80‖4 = 33.33 + 3.81 = 37.14 Ω ·. . 37.14‖20 = (37.14)(20) 57.14 = 13 Ω ·. . Rab = 13 + 13 + 7 = 33 Ω docsity.com 3–40 CHAPTER 3. Simple Resistive Circuits R6 = (60)(80) 200 = 24 Ω Now redraw the circuit using the wye equivalents. Rab = 1.5 + 12.5 + (25 + 71 + 24)‖(25 + 31 + 24) + 18 = 1.5 + 12.5 + (120‖85) + 18 = 1.5 + 12.5 + 48 + 18 = 80 Ω [b] When vab = 400 V ig = 400 80 = 5 A io = 120 120 + 80 (5) = 3 A p31Ω = (31)(3)2 = 279 W P 3.59 [a] After the 20 Ω—80 Ω—40 Ω wye is replaced by its equivalent delta, the circuit reduces to docsity.com Problems 3–41 Now the circuit can be reduced to Req = 44 + 280‖92.5 = 113.53 Ω ig = 5/113.53 = 44.04 mA i = (280/372.5)(44) = 33.11 mA v52.5Ω = (52.5)(33.11 m) = 1.74 V io = 1.74/210 = 8.28 mA [b] v40Ω = (40)(33.11 m) = 1.32 V i1 = 1.32/56 = 23.65 mA [c] Now that io and i1 are known return to the original circuit i80Ω = 44.04 m − 23.65 m = 20.39 mA i20Ω = 23.65 m − 8.28 m = 15.37 mA i2 = i80Ω + i20Ω = 35.76 mA [d] pdel = (5)(44.04 m) = 220.2 mW docsity.com 3–42 CHAPTER 3. Simple Resistive Circuits P 3.60 [a] After the 30 Ω—60 Ω—10 Ω delta is replaced by its equivalent wye, the circuit reduces to Use current division to calculate i1: i1 = 22 + 18 22 + 18 + 4 + 6 (5 A) = 40 50 (5 A) = 4 A [b] Return to the original circuit and write a KVL equation around the upper left loop: (22 Ω)i22Ω + v − (4 Ω)(i1) = 0 so v = (4 Ω)(4 A) − (22 Ω)(5 A − 4 A) = −6 V [c] Write a KCL equation at the lower center node of the original circuit: i2 = i1 + v 60 = 4 + −6 60 = 3.9 A [d] Write a KVL equation around the bottom loop of the original circuit: −v5A + (4 Ω)(4 A) + (10 Ω)(3.9 A) + (1 Ω)(5 A) = 0 So, v5A = (4)(4) + (10)(3.9) + (1)(5) = 60 V Thus, p5A = (5 A)(60 V) = 300 W P 3.61 Subtracting Eq. 3.42 from Eq. 3.43 gives R1 − R2 = (RcRb − RcRa)/(Ra + Rb + Rc). Adding this expression to Eq. 3.41 and solving for R1 gives R1 = RcRb/(Ra + Rb + Rc). To find R2, subtract Eq. 3.43 from Eq. 3.41 and add this result to Eq. 3.42. To find R3, subtract Eq. 3.41 from Eq. 3.42 and add this result to Eq. 3.43. Using the hint, Eq. 3.43 becomes R1 + R3 = Rb[(R2/R3)Rb + (R2/R1)Rb] (R2/R1)Rb + Rb + (R2/R3)Rb = Rb(R1 + R3)R2 (R1R2 + R2R3 + R3R1) docsity.com Problems 3–45 [b] When R = RL, the circuit reduces to io = ii(3RL) 4.5RL = 1 1.5 ii = 1 1.5 vi RL , vo = 0.75RLio = 1 2 vi, Therefore vo vi = 0.5 P 3.65 [a] 3.5(3R − RL) = 3R + RL 10.5R − 1050 = 3R + 300 7.5R = 1350, R = 180 Ω R2 = 2(180)(300)2 3(180)2 − (300)2 = 4500 Ω [b] vo = vi 3.5 = 42 3.5 = 12 V io = 12 300 = 40 mA i1 = 42 − 12 4500 = 30 4500 = 6.67 mA ig = 42 300 = 140 mA i2 = 140 m − 6.67 m = 133.33 mA i3 = 40 m − 6.67 m = 33.33 mA i4 = 133.33 m − 33.33 m = 100 mA docsity.com 3–46 CHAPTER 3. Simple Resistive Circuits p4500 top = (6.67 × 10−3)2(4500) = 0.2 W p180 left = (133.33 × 10−3)2(180) = 3.2 W p180 right = (33.33 × 10−3)2(180) = 0.2 W p180 vertical = (100 × 10−3)2(180) = 1.8 W p300 load = (40 × 10−3)2(300) = 0.48 W The 180 Ω resistor carrying i2 dissipates the most power. [c] p180 left = 3.2 W [d] Two resistors dissipate minimum power – the 4500 Ω and the 180 Ω carrying i3. [e] Both resistors dissipate 0.2 W or 200 mW. P 3.66 [a] va = vinR4 Ro + R4 + ∆R vb = R3 R2 + R3 vin vo = va − vb = R4vin Ro + R4 + ∆R − R3 R2 + R3 vin When the bridge is balanced, R4 Ro + R4 vin = R3 R2 + R3 vin ·. . R4 Ro + R4 = R3 R2 + R3 Thus, vo = R4vin Ro + R4 + ∆R − R4vin Ro + R4 = R4vin [ 1 Ro + R4 + ∆R − 1 Ro + R4 ] = R4vin(−∆R) (Ro + R4 + ∆R)(Ro + R4) ≈ −(∆R)R4vin (Ro + R4)2 docsity.com Problems 3–47 [b] ∆R = 0.03Ro Ro = R2R4 R3 = (1000)(5000) 500 = 10,000 Ω ∆R = (0.03)(104) = 300 Ω ·. . vo ≈ −300(5000)(6)(15,000)2 = −40 mV [c] vo = −(∆R)R4vin (Ro + R4 + ∆R)(Ro + R4) = −300(5000)(6) (15,300)(15,000) = −39.2157 mV P 3.67 [a] approx value = −(∆R)R4vin (Ro + R4)2 true value = −(∆R)R4vin (Ro + R4 + ∆R)(Ro + R4) ·. . approx value true value = (Ro + R4 + ∆R) (Ro + R4) ·. . % error = [ Ro + R4 + ∆R Ro + R4 − 1 ] × 100 = ∆R Ro + R4 × 100 But Ro = R2R4 R3 ·. . % error = R3∆R R4(R2 + R3) [b] % error = (500)(300) (5000)(1500) × 100 = 2% P 3.68 ∆R(R3)(100) (R2 + R3)R4 = 0.5 ∆R(500)(100) (1500)(5000) = 0.5 ·. . ∆R = 75 Ω % change = 75 10,000 × 100 = 0.75% docsity.com 3–50 CHAPTER 3. Simple Resistive Circuits R2 = (1 + 2σ)2R1 = 1.1435 Ω Using symmetry, R4 = R2 = 1.1435 Ω R5 = R1 = 1.0372 Ω Rc = Rb = 0.0068 Ω Rd = Ra = 0.0259 Ω Test the calculations by checking the power dissipated, which should be 120 W/m. Calculate D, then use Eqs. (3.58)-(3.60) to calculate ib, i1, and i2: D = (R1 + 2Ra)(R2 + 2Rb) + 2R2Rb = 1.2758 ib = Vdc(R1 + R2 + 2Ra) D = 21 A i1 = VdcR2 D = 10.7561 A i2 = Vdc(R1 + 2Ra) D = 10.2439 A It follows that i2bRb = 3 W and the power dissipation per meter is 3/0.025 = 120 W/m. The value of i21R1 = 120 W/m. The value of i 2 2R2 = 120 W/m. Finally, i21Ra = 3 W/m. P 3.72 From the solution to Problem 3.71 we have ib = 21 A and i3 = 10 A. By symmetry ic = 21 A thus the total current supplied by the 12 V source is 21 + 21 + 10 or 52 A. Therefore the total power delivered by the source is p12 V (del) = (12)(52) = 624 W. We also have from the solution that pa = pb = pc = pd = 3 W. Therefore the total power delivered to the vertical resistors is pV = (8)(3) = 24 W. The total power delivered to the five horizontal resistors is pH = 5(120) = 600 W. ·. . ∑pdiss = pH + pV = 624 W = ∑pdel P 3.73 [a] σ = 0.03/1.5 = 0.02 Since the power dissipation is 150 W/m the power dissipated in R3 must be 200(1.5) or 300 W. Therefore R3 = 122 300 = 0.48 Ω From Table 3.1 we have R1 = (1 + σ)2R3 (1 + 2σ)4 = 0.4269 Ω Ra = σR1 = 0.0085 Ω R2 = (1 + 2σ)2R1 = 0.4617 Ω docsity.com Problems 3–51 Rb = (1 + 2σ)2σR1 4(1 + σ)2 = 0.0022 Ω Therefore R4 = R2 = 0.4617 Ω R5 = R1 = 0.4269 Ω Rc = Rb = 0.0022 Ω Rd = Ra = 0.0085 Ω [b] D = [0.4269 + 2(0.0085)][0.4617 + 2(0.0022)] + 2(0.4617)(0.0022) = 0.2090 i1 = VdcR2 D = 26.51 A i21R1 = 300 W or 200 W/m i2 = R1 + 2Ra D Vdc = 25.49 A i22R2 = 300 W or 200 W/m i21Ra = 6 W or 200 W/m ib = R1 + R2 + 2Ra D Vdc = 52 A i2bRb = 6 W or 200 W/m isource = 52 + 52 + 12 0.48 = 129 A pdel = 12(129) = 1548 W pH = 5(300) = 1500 W pV = 8(6) = 48 W∑ pdel = ∑ pdiss = 1548 W docsity.com