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JEE2300 PSpice Tutorial 2 Problem 1. Use a Y-to- transformation to find (a) io; (b) i1; (c) i2; and (d) the power delivered by the ideal current source in the circuit show below. Use PSpice to confirm your work. 320 20 i1 50 io i2 240 1A 100 600 Figure 1. (a) Need to mark Y network to perform Y-to- transformation. That is done below: 320 a 1A R1 = 20 i20 i50 i1 R2 = 50 i240 i2 240 R3 =100 c Figure 2. 1 600 b io Now we perform Y-to- transformation between nodes a,b, and c: R1 R2 R2 R3 R3 R1 1000 5000 2000 8000 400 R1 20 20 R R R2 R3 R3 R1 1000 5000 2000 8000 Rb 1 2 160 R2 50 50 R R R2 R3 R3 R1 1000 5000 2000 8000 Rc 1 2 80 R3 100 100 Ra Now the circuit looks like this: 320 i240 1A iab R = 80 c a iac 240 Rb = 160 c Figure 3. Using KVL from Figure 3 we can see that: i240 240 iac 160 (1) ibc 400 io 600 (2) i1 320 iab 80 (3) We can simplify this circuit doing the following: Req1 = (240) || Rb = 96 and i96 = i240 + iac (4) 2 Ra = 400 i1 b ibc io 600 Req2 = (600) || Ra = 240and i64 = ibc + io (5) Req3 = (320) || Rc = 64and i64 = iab + i1 (6) New circuit looks like this: a i64 Req3 = 64 b i96 1A Req1 = 96 c Figure 4. From Figure 4 using KCL: 1A i64 i96 Also using KVL for the Figure 4: i64 64 240 i96 96 From the last two equations we get i64 0.24 A i96 0.76 A Using (2) and (5) we get: ibc 144mA io 96mA 3 Req2 = 240 (b) From (3) and (6) we get: iab 192mA i1 48mA (c) From (1) and (4) we get iac 456mA i240 304mA Using KCL from Figure 2 we have: 1A = i240 + i20 + i1 i20 = 648mA. Also using KCL from Figure 2 we have: i50 = io - i1 i50 = 48mA ,and using KCL again from Figure 2 we have: i20 = i50 + i2 i2 = 600mA. (d) From Figure (2) we have that P1A = - (1A)v240 P1A = - (1A)(240)(i240) = - (1A)(240)(304mA) = - 72.96W – power delivered by the 1A ideal current source P240 = (i240)2(240) = (304mA)2 (240) = 22.17984W P20 = (i20)2(20) = (648mA)2 (20) = 8.39808W P320= (i1)2 (320) = (48mA)2 (320) = 0.73728W + P100= (i2)2 (100) = (600mA)2 (100) = 36W P50= (i50)2 (50) = (48mA)2 (50) = 0.1152W P600= (io)2 (600) = (96mA)2 (600) = 5.5296W 72.96W – total power absorbed by the resistive network in the circuit 4 Problem 2. Use the node-voltage method to find v1 and v2 in the circuit shown in Figure 5. 25 1 + 2.4A v1 i25 2 i125 i250 250 125 i375 + 375 v2 - 3.2A 3 Figure 5. There are 3 essential nodes in the circuit from Figure 5. Hence n = 3 and we need (n-1) = 2 equations to determine all voltages between essential nodes. In this circuit there are 2 voltages like that and we marked them as v1 and v2. Essential node 3 is common node and we marked it like that. We also chose current directions for each branch. Using node voltage method we can write two independent equations: v v v Node 1: 2.4 A 1 2 1 0 125 25 v2 v v v 2 2 1 0 375 250 25 We have two equations with two unknown voltages v1 and v2. If we solve these two equations we get: v1 = 25V v2 = 90V. Node 2: 3.2 A Problem 3. (a) Use the mesh-current method to find the branch currents ia,ib, and ic in the circuit Figure 6. (b) Repeat (a) if the polarity of the 64V source is reversed. ia 3 ic 4 ib + 40V - I 45 2 II 1.5 Figure 6. 5 64V + (a) Number of essential nodes is ne = 2. Number of essential branches with unknown currents is be = 3. We can write be – (ne -1) = 3 – 1 = 2 independent mesh-current method equations. Mesh I: (-40V) + (3)iI + (45)(iI – iII) + (2)iI = 0 Mesh II: (-64V) + (4)iII + (45)(iII – iI) + (1.5)iII = 0 ____________________________________________________________________________ We have two equations with two unknown mesh currents iI and iII and when we solve them: iI = 9.8A iII = 10A. From here we have that ia = iI = 9.8A ib = iI – iII = -0.2A ic = - iII = -10A. (b) If 64V independent voltage source has reversed polarity then we have the following mesh equations: Mesh I: (-40V) + (3)iI + (45)(iI – iII) + (2)iI = 0 Mesh II: (64V) + (4)iII + (45)(iII – iI) + (1.5)iII = 0 _____________________________________________________________________________ We have two equations with two unknown mesh currents iI and iII and when we solve them: iI = -1.72A iII = - 2.8A. From here we have that ia = iI = -1.72A ib = iI – iII = 1.08A ic = - iII = 2.8A. Problem 4. (a) Use the mesh-current method to find how much power the 12A current source delivers to the circuit shown in Figure 7. (b) Find the total power delivered to the circuit. (c) Check your calculations by showing that the total power developed in the circuit equals the total power dissipated. 6 10 + 600V - 8 2 I 40 14 + 400V - II 2 1 4 3 + v14 - 12A Figure 7. (a) Number of essential nodes is ne = 4. Number of essential branches with unknown currents is be = 5. We can write be – (ne -1) = 5 – 3 = 2 independent mesh-current method equations. Mesh I: (-600V) + (10)iI + (40)(iI – iII) + (14)(iI +12A) = 0 Mesh II: (400V) + (8)iII + (40)(iII – iI) + (2)(iII +12A) = 0 __________________________________________________________________________ We have two equations with two unknown mesh currents iI and iII and when we solve them: iI = 2.9A iII = -6.16A. Therefore the power delivered by the 12A current source is: P12A = (v14)(12A) P12A = (v13+v34)(12A) P12A = (v13+v34)(12A) P12A = (v13+v34)(12A) P12A = [-(14)(12A+iI)-(2)(12A+iII)](12A) P12A = [-(14)(12A+2.9A)-(2)(12A-6.16A)](12A) P12A = [-(14)(14.9A)-(2)(5.84A)](12A) P12A = - 2643.36W. 7 (b) P12A = - 2643.36W P600V = - (600V)iI = - 1740W + P400V = (400V)iII = - 2464W ________________________________________________________ PTOTAL DEVELOPED = - 6847.36W (c) P10 = (iI)2(10) = (2.9A)2(10) = 84.1W 2 2 P8= (iII) (8) = (-6.16A) (8) = 303.5648W 2 2 P2= (iII+12A) (2) = (-6.16A +12A) (2) = 68.2112W + P40= (iI – iII)2(40) = (2.9A + 6.16A)2(40) = 3283.344W P14= (iI+12A)2(14) = (2.9A + 12A)2(14) = 3108.14W ________________________________________________________ PTOTAL ABSORBED = 6847.36W Problem 5. (a) Use the principle of superposition to find the voltage v in the circuit shown in Figure 8. (b) Find the power dissipated in the 20 resistor. 6A 5 1 8 2 3 + + - 75V v 20 4 Figure 8. 8 12 (a) First we analyze circuit without 6A independent current source. The new circuit will look like this: 5 1 i 1' 2 i 3' 8 3 i 2' + - 75V 20 12 4 Figure 9. Using KCL at node 2 from Figure 9 we have: i1 ’ = i2 ’ + i3 ’ (1) Using KVL from Figure 9 we have: (20)i2’ = (8+12)i3’ i2 ’ = i3 ’ (2) Also using KVL we have: 75V = (5)i1’ + (20)i2’ (3) Using (1), (2) and (3) we get: i1’ = 5A i2’ = 2.5A i3’ = 2.5A. Second we analyze circuit without 75V independent voltage source. The new circuit will look like this: 9 6A 5 1 i 1" 2 i4 i 3" 20 12 Figure 10. Using KVL from Figure 10 we can see that (20)i2” = - (5i1” (4) Using KCL from Figure 10 we can see that at node 1 we have 6 + i2 ” + i 3 ” = i1 ” (5) Combining (4) and (5) we have 6 + 5i2” + i3” = 0 (6) Using KCL from Figure 10 at node 3 6 - i4 + i3” = 0 (7) Using KVL from Figure 10 we have that (20)i2” = (8i4 + (12 i3” (8) Combining (7) and (8) we get 20i2” - 20 i3” = 48 3 i 2" 4 i1” = - 4i2” 8 (9) From (6) and (9) we have: 10 6 + 5i2” + i3” = 0 20i2” - 20 i3” = 48 ________________ 5i2” + i3” = -6 20i2” - 20 i3” = 48 ________________ From here i2” = -0.6A If we add both i2’ and i2” we get total current i2 determined by the superposition method: i2 = i2’ + i2” = 2.5 -0.6 = 1.9A. And voltage v is: v = (20)i2 = 38V. (b) The power dissipated in the 20 resistor is v 2 38 72.2W 20 20 2 P20 Problem 6. The Inverting-Amplifier Circuit Design The Inverting-Amplifier Circuit Design is shown in Figure 11. 10k +15V 1k + - vs + -15V Figure 11. 11 + vo - 10k vo v s 10v s 1k (a) If vs = 1Vdc then vo = -10Vdc (the op amp is operating in linear region). (b) If vs = 2Vdc then vo = -15Vdc (the op amp is operating in negative saturation region). For the Inverting-Amplifier Circuit Design shown in Figure 12, 10k +15V 1k + vs + -15V + vo - Figure 12. 10 k vo v s 10v s 1k (c) If vs = 1Vdc then vo = 10Vdc (the op amp is operating in linear region). (d) If vs = 2Vdc then vo = 15Vdc (the op amp is operating in positive saturation region). Problem 7. The Summing-Amplifier Circuit Design The Summing-Amplifier Circuit Design is shown in Figure 13. Rf Ra + v a - +15V Rb Rc + v - b + v - c + -15V Figure 13. 12 + vo - Rf Rf Rf vo va vb vc R R R b c a (a) If Rf = Ra = Rb = Rc = 1k, va = 1V, vb = 2V, and vc = 3V, then vo = -6V (the op amp is operating in linear region). (b) If Rf = Ra = Rb = Rc = 1k, va = 5V, vb = 5V, and vc = 7V, then vo = -15V (the op amp is operating in negative saturation region). Problem 8. The Non-Inverting-Amplifier Circuit Design Rf Rs +10V Rg + -10V + - + vo - vg Figure 14. For the Non-Inverting-Amplifier Circuit Design shown in Figure 14 Rs R f vo Rs v g (a) If vg = 2Vdc, Rs = Rg = 1k, and Rf = 2k, then vo = 6Vdc (the op amp is operating in linear region). (b) If vg = 5Vdc, Rs = Rg = 1k, and Rf = 2k, then vo = 10Vdc (the op amp is operating in positive saturation region). 13