Question

Mini-Prj 2. Extraction of Thevenin and Norton Equivalent Circuits by LTspice Part A. Wheatstone Bridge Circuit with a Voltage Source Vs R5 R1 R2 Vs RL R3 R4 For the circuit as shown below, given that R1= 9 Ω, R2= 17 Ω, R3= 9 Ω' R,-18 Ω, R5= 19 Ω , RL= 2 Ω ,V,-74 V I. Wheatstone Bridge Circuit Analysis (a) Determining the load voltage Vi-Vab for the Wheatstone bridge circuit with LTspice Submit Answer Tries 0/3 (b) Determining the load current Iu following from a to b for the Wheatstone bridge circuit with LTspice. Submit Answer Tries 0/3 II. Extraction of Thevenin Equivalent Circuit for this Wheatstone Bridge Circuit RTh VTh RL (c) Determining Thevenin equivalent voltage VTh for this Wheatstone bridge circuit, where VTh is extracted from the open circuit voltage at the terminals a, b without the load, RL Submit Answer Tries o/3 (d) As shown below, extracting Thevenin equivalent resistance RTh for this Wheatstone bridge circuit by setting Vs-0 V and adding an external current source with Le-1.2 A. Thus, Thevenin equivalent resistance can be evaluated via RThV (Question: why not use Vab instead?) R5 R1 R2 Ie Vs 0 R3 R4 (S2) Submit Answer Tries 0/3 (e) With the extracted VTh and RTh and using Thevenin equivalent circuit, recalculate the load voltage Vu-VLTh-Vab for the Wheatstone bridge circuit with LTspice VLTh Submit Answer Tries 0/3 ) With the extracted VTh and RTh and using Thevenin equivalent circuit, recalculate the load current IL-ILTh for the Wheatstone bridge circuit with LTspice. ILM" Submit Answer Tries 0/3 III. Extraction of Norton Equivalent Circuit for this Wheatstone Bridge Circuit IN RN RL (g) Determining Norton equivalent current IN for this Wheatstone bridge circuit, where IN is extracted from the source transformation IN VTh/RTh IN- Submit Answer Tries 0/3 (h) Determining Norton equivalent resistance RN, where RN is extracted from the source transformation RN-RTh RN Submit Answer Tries 0/3 (i) with the extracted IN and RN and using Norton equivalent circuit, recalculate the load voltage VL-VLN=Vab for the Wheatstone bridge circuit with LTSpice. VLN Submit Answer Tries 0/3 ) With the extracted IN and RN and using Norton equivalent circuit, recalculate the load current IL ILN for the Wheatstone bridge circuit with LTspice. Submit Answer Tries 0/3 ILN Mini-Prj 2. Extraction of Thevenin and Norton Equivalent Circuits by LTspice Part B. Wheatstone Bridge Circuit with a Current Source Is R5 R1 R2 Is RL R3 R4 For the circuit as shown below, given that R1-22 Ω, R2-15 Ω, R3-28 Ω, R4-9 Ω, R5-29 Ω , R.-16 Ω, 1,-6 A. I. Wheatstone Bridge Circuit Analysis (a) Determining the load voltage VL-Vab for the Wheatstone bridge circuit with LTspice. Submit Answer Tries 0/4 (b) Determining the load current IL following from a to b for the Wheatstone bridge circuit with LTspice Submit Answer Tries 0/4 II. Extraction of Thevenin Equivalent Circuit for this Wheatstone Bridge Circuit RTh VTh RL (c) Determining Thevenin equivalent voltage VTh for this Wheatstone bridge circuit, where VTh is extracted from the open circuit voltage at the terminals a, b without the load, RL Submit Answer Tries 0/3 d) As shown below, extracting Thevenin equivalent resistance RTh or this wheatstone bridge circuit by setting 1s=0 A and adding an external current source with e=1.2 A. Thus, Thevenin equivalent resistan can be evaluated via RTh Va/e (Question: why not use Vab instead?) R5 R1 R2 Is Ie R3 R4 0 Submit Answer Tries 0/3 (e) With the extracted VTh and RTh and using Thevenin equivalent circuit, recalculate the load voltage V VLTh-Vab for the Wheatstone bridge circuit with LTspice. Submit Answer Tries 0/3 (f) With the extracted VTh and RTh and using Thevenin equivalent circuit, recalculate the load current ILILTh following from left to right for the Wheatstone bridge circuit with LTspice ILTh Submit Answer Tries 0/3 III. Extraction of Norton Equivalent Circuit for this Wheatstone Bridge Circuit IN RN RL (g) Determining Norton equivalent current IN for this wheatstone bridge circuit, where IN įs extracted from the source transformation 1N= Vrh/RT- 1 (A) Submit Answer Tries 0/3 (h) Determining Norton equivalent resistance RN, where RN is extracted from the source transformation RN RN- RTh Submit Answer Tries 0/3 ) With the extracted IN and RN and using Norton equivalent circuit, recalculate the load voltage VL-VLN-Vab for the Wheatstone bridge circuit with LTspice VUN Submit Answer Tries 0/3 ) With the extracted IN and Rv and using Norton equivalent circuit, recalculate the load current IL ILN following from left to right for the Wheatstone bridge circuit with LTspice. ILN Submit Answer Tries 0/3

Answer 1

I. I have simulated the given wheat stone bridge using LTspice and results are shown below

LTspice XVII Draft1 Eile View Plot Settings Simulation Iools Window Help Egi Adi刍ㄧㄥ女叩之 3 D-纱屮り EmE] Aa "op Draftt R5 19 R2 17 Vs RL 74 V R3 R4 18 .trán 1 Draft1 14.373V 4.370V 14.367V 14.364V 14.361V 14.358V 14.355V 14.352V 14.349V 14.346V V(a) 14.343V 15ps 18ps 21ps 24ps 27ps 30ps Ops 3ps 6ps 9ps 12ps

LTspice XVII - Draft Eile View Plot Settings Simulation Iools Window Help ] Τ 茅49 Q Q皋賒ig E 義더 R龉生刍 / 回ぐ 3 D-ep'Y Of E Aa.op Draft1 RS 19 R2. 17 9 Vs RL 74 V R3 R4 9 18 .tran 1 Draft V(b) 14.427V 14.424V 14.421V 14.418V 4.415V 14.412V 4.409V 14.406V 14.403V 14.400V 14.397V 0.0fs 0.4fs 0.8fs 1.2fs 1.6fs 2.0fs 2.4fs 2.8fs 3.2fs 3.6fs 4.0fs 4.4fs 4.8fs

Waveforms for V(a) and V(b) can be shown on the same graph,but I have shown them on different graphs in this first step.

From simulations we have, V(a)=14.358 V and V(b)=14.411 V therefore V(ab)=V(a)-V(b)=14.358-14.411 = -0.052 V

Therefore, V(ab) = V_L= -0.052 V.

LTspice XVII - Draftl File View Plot Settings Simulation Iools Window Help Draftt Draft R5 19 R2 Vs RL 74 V R3 R4 18 .tran 1 Draft 26.703mA 6.700mA 26.697mA (RI) 6.694mA 26.691mA 26.688mA 26.685mA 26.682mA 26.679mA 6.676mA 6.673mA 0.0000fs 0.0001fs 0.0002fs 0.0003fs 0.0004fs 0.0005fs 0.0006fs 0.0007fs 0.0008fs 0.0009fs 0.0010fs 0.0011fs 0.0012fs 0.0013fs 0.0014fs

From simulation graph .Load current I(RL)=26.688 mA

Therefore, I(RL)=I_L=0.0268 A

2. Now finding Vth by removing R_L , we have

LTspice XVII - Draftl File View Plot Settings Simulation Iools Window Help Drat1 Draft1 Draft1 R5 19 R1, R2 9 Vs 74 V R4 18 R3 9 .trăn 1 Draft 14.68V 14.64V 14.60V 14.56V 14.52V 14.48V 14.44V 14.40V 14.36V 14.32V 14.28V 14.24V 14.20V V(a) V(b) 0.0s 0.1s 0.2s 0.3s 0.4s 0.5s 0.6s 0.7s 0.8s 0.9s 1.0s

V_a(black line in graph)=14.24 V, V_b(Pink line in graph)=14.64 V

Therefore , Vth= V_a - V_b = 14.24 - 14.64 = -0.4 V

Therefore Vth= -0.4 V

For Rth, current source=1.2 A is applied between a and b as shown below

Therefore V_ba=V_b - V_a = 25 - 9 = 16 V , Current source (I1)=1.2 A

LTspice XVII Draft1 Eile View Plot Settings Simulation Iools Window Help LDrDraft Draft1 R5 19 R1 R2 17 9 Vs 74 V 1.2 R4 18 R3 9 .tran 1 Draftl V(a) V(b) 4V 0V 18V 16V 14V 12V 10V 8V 0.0s 0.1s 0.2s 0.3s 0.4s 0.5s 0.6s 0.7s 0.8s 0.9s 1.0s

Rth=V_ba/I1 = 16/1.2 = 13.33 ohm

Therefore , Rth=13.33 ohm

Therefore Thevenin equivalent circuit we have ,

V_b=0 V(gnd), V_a(black line)= -52.2 mV= -0.052 V , therefore V_a - V_b= V_ab=V_Lth= -0.052 V ,Hence same as above calculated.

LTspice XVII - Draft2 File View Plot Settings Simulation Tools Window Help Draft22 raft2 Rth 13.33 Vth RL 2 0.4 .tran 1 V(a) (RI) 51.0mV 1.2mV 1.4mV 1.6mV 1.8mV 52.0mV 52.2mV 52.4mV 52.6mV 52.8mV 53.0mV 63.2mV -25.0mA -25.2mA -25.4mA -25.6mA 25.8mA -26.0mA -26.2mA 26.4mA -26.6mA -26.8mA -27.0mA -27.2mA 0.0s 0.1s 0.2s 0.3s 0.4s 0.5s 0.6s 0.7s 0.8s 0.9s 1.0s

I_L= 26.2 mA(pink line in graph)

Hence I_Lth= 0.0262 A (nearly same as above calculated)

Now Norton equivalent circuit we have , I_n=Vth/Rth= -0.4/13.33 = -0.03 A , R_n=Rth= 13.33 ohm

LTspice XVII - Draft3 File View Plot Settings Simulation Tools Window Help DrafDraDraf2Draft2DratDraft? Draft3 In RL 13.33 2 0.03 .tran 1 Draft3 51.0mV 51.2mV 51.4mV 51.6mV 51.8mV 52.0mV 52.2mV 52.4mV 52.6mV 52.8mV 53.0mV 53.2mV V(a) (RI) 25.0mA 25.2mA 25.4mA -25.6mA -25.8mA -26.0mA 26.2mA 26.4mA 26.6mA -26.8mA -27.0mA -27.2mA 0.0s 0.1s 0.2s 0.3s 0.4s 0.5s 0.6s 0.7s 0.8s 0.9s 1.0s

Therefore from graph we have V_Ln=V_a - V_b= -52.2mV -0 = -52.2 mV , therefore VLn= - 0.052 V

I_Ln=0.0262 A

As per HomeworkLib policy I have solved first question as you have posted multiple question. feel free to ask if you have any query.Thank you