ENGINEERING-43
Thévenin & Norton Equivalents
Lab-08
Lab Data Sheet – ENGR-43 Lab-08
Lab Logistics
Experimenter: Bruce Mayer, PE
Recorder:
Date: Mar06
Equipment Used (maker, model, and serial no. if available)
Fluke 8050A DMM; S/N 4630210
Tektronix PS280 DC Power Supply; S/N TW527171
Special Note
This laboratory exercise entails a significant amount of circuit construction &
measurement effort/time. For this reason:
 Please COMPLETE ALL MEASUREMENTS in Table I, Table II, Table III, and
Table VII BEFORE completing any of the calculations
Directions
1. Check out a DMM and Power/Probe Leads for the Power-Supply and DMM
2. Go to the side counter, collect resistors, “bread board”, and leads required to construct the
circuit shown in Figure 1. Configure the Power Supply Outputs to the INDEPENDENT
mode as indicated in Figure 2.
3. Make the Measurement and Calculations needed to complete Table I, Table II, Table III,
Table IV, Table V, and Table VI.
© Bruce Mayer, PE • Chabot College • 282219678 • Page 1
Figure 1 - Connection Diagram for the Dual Voltage Supply Experimental Network.
Vs1 = 13V (nominal), Vs2 = 5V (nominal), R1 = 1.4-2.4 kΩ, R2 = 2.9-5.1 kΩ, R3 = 6.5-8.8
kΩ, R4 = 9.5-16 kΩ, RL = 2.6-3.6 kΩ (3.3 kΩ nominal)
© Bruce Mayer, PE • Chabot College • 282219678 • Page 2
Vs1 = 13.0
Vs2 = 5.0
OUT
OUT
-
Vs1
+
13V
-
Vs2
+
5V
Figure 2 - Connection Diagram for the Dual Voltage Supplies used to Power the
Experimental Network of Figure 1. Set Vs1 to 13V (nominal), Vs2 to 5V (nominal).
Table I – Component Actual-Values by DMM Measurement
Vs1 =
Vs2 =
R1 =
R2 =
R3 =
R4 =
RL =
13.035 V
5.009 V
2.089 kΩ
3.803 kΩ
7.584 kΩ
11.472 kΩ
3.478 kΩ
© Bruce Mayer, PE • Chabot College • 282219678 • Page 3
Table II – Node Voltages: Calculations & DMM-Measurements
Value
Determination
V1
V2
V3
Calculated
8.368 V
13.035 V
3.456 V
Measured
8.466 V
13.057 V
3.459 V
%
1.17%
0.0153%
0.0868%


Calculate Node Voltages using the component actual-values from Table I
%J = 100x(VJ,meas – VJ,calc)/VJ,calc
Table III – Branch Currents: Calculations & DMM-Measurements
Value
Determin.
I1 (mA)
I2 (mA)
I3 (mA)
I4 (mA)
Is1 (mA)
Is2 (mA)
IL (mA)
Calculated
-2.234
3.428
-0.456
0.729
-5.662
1.505
1.049
Measured
-2.182
3.419
-0.453
0.733
-5.573
1.442
0.992
%
2.33%
0.263%
0.658%
0.549%
1.57%
4.19%
5.43%


Calculate Branch Currents using the component actual-values from Table I
%J = 100x(IJ,meas – IJ,calc)/IJ,calc
Table IV – Power Absorbed by Voltage Supplies: Component and VI Calculations
Value
Calculations
PVs1
PVs2
ΣPVsj
Component: Calc1
-73.804 mW
7.539 mW
66.265 mW
Measured VI: Calc2
-72.655 mW
7.223 mW
65.432 mW
%
1.557%
4.192%
1.257%




For ALL power calculations assume that the PASSIVE Sign convention relates component
voltage-polarities and current-directions
Calculation-1: Use the component actual-values from Table I
Calculation-2: Use the measured values of V-across, and I-thru for the two supplies from
Table II and Table III
%J = 100x(PJ,meas – PJ,calc)/PJ,calc
© Bruce Mayer, PE • Chabot College • 282219678 • Page 4
Table V – Power Absorbed by Resistors: Component and VI Calculations
Value
Calcs
PR1 (mW)
PR2 (mW)
PR3 (mW)
PR4 (mW)
PRL (mW)
ΣPRj (mW)
Component:
Calc1
10.426
44.69
1.577
6.097
3.827
66.617
Measured
VI: Calc2
9.946
44.455
1.556
6.164
3.423
65.544
%
4.603%
0.526%
1.33%
1.099%
10.56%
1.611%




For ALL power calculations assume that the PASSIVE Sign convention relates component
voltage-polarities and current-directions
Calculation-1: Use the component actual-values from Table I
Calculation-2: Use the measured values of V-across, and I-thru for the two supplies from
Table II and Table III
%J = 100x(PJ,meas – PJ,calc)/PJ,calc
Table VI – Power Balance
Value
Calculations
ΣPVsj
ΣPRj
% Out of Balance, ΔOB%
Component: Calc1
-66.265 mW
66.617 mW
0.265%
Measured VI: Calc2
-65.432 mW
65.544 mW
0.0855%



Calculation-1: Use the component-calculations from Table IV and Table V
Calculation-2: Use the measured VI calculations from Table IV and Table V
OB% by this Equation:
 OB % 
P
P
Vsj
  PRj
Vsj

P
Rj
© Bruce Mayer, PE • Chabot College • 282219678 • Page 5
Directions (continued)
4. Remove the Load Resistor, RL, from the circuit to leave OPEN that branch of the circuit as
shown in Figure 3
5. Make the measurements and calculations need to complete Table VII
Figure 3 - Connection Diagram for the Thevenin Equivalent Circuit. Use the Same
Supply-Voltages and Resistor-Values as used to make previous measurements.
© Bruce Mayer, PE • Chabot College • 282219678 • Page 6
Table VII – Thevenin Component Determination
Value
4.884 V
Quantity, and Determination-Method
= Voc By DMM Measurment
3.383 mA
= Isc By DMM Measurment
 Hint: The DMM itself acts as the Short Circuit
1.444 kΩ
= RTH,VI = Voc/Isc
 Use DMM Measured Voc and Isc
1.4317 kΩ
= RTH,SD by Source Deactivation
 DeActivate the Voltage Sources by REMOVING them from the Ckt,
and REPLACING them with a wire
 Measure using the DMM the resulting Resistance as seen from the
Voc terminals
1.4379 kΩ
= RTH,avg = [RTH,VI + RTH,SD]/2
Directions (continued)
6. In the Space Below Neatly Draw the THEVENIN Equivalent for the Circuit shown in Figure
3 with the Load Resistor REATTACHED. Use RTH,avg from Table VII.
7. For the Thevenin Equivalent Circuit calculate the Load Current, IL, and mark its magnitude
and direction on the circuit Diagram.
© Bruce Mayer, PE • Chabot College • 282219678 • Page 7
1.438k
3.478k
RL
4.884 V +
- Vth
ILTH  0.9935mA
Directions (continued)
8. In the Space Below Neatly Draw the NORTON Equivalent for the Circuit shown in Figure 3
with the Load Resistor REATTACHED. Use RTH,avg from Table VII.
9. For the NORTON Equivalent Circuit calculate the Load Current, IL, and mark its magnitude
and direction on the circuit Diagram.
© Bruce Mayer, PE • Chabot College • 282219678 • Page 8
ILN  0.9896mA
Directions (continued)
10. Return all lab hardware to the “as-found” condition
Directions (continued)
11. QUESTION: How does IL by the Thevenin and Norton Equivalents compare to the DMMMeasured value from Table III?
Using the Single-Loop-Circuit method on the Thevenin form find IL as
ILTH 
4.884V
 0.9935mA
1.438  3.478k
Using the Current-Divider method on the Norton form find IL as
© Bruce Mayer, PE • Chabot College • 282219678 • Page 9
1.438k
ILN  3.383mA
 0.9896mA
1.438  3.478k
From Table III note the measured Load Current of 0.992 mA. Using the MEASURED Value as
the BASELINE, find the Δ% for both equivalent Circuits
0.9935  0.992
Th % 
 0.00151  0.151%
0.992
0.9896  0.992
N % 
 0.00242  0.242%
0.992
The Thevenin and Norton equivalents provide estimated Load Currents that are within less
than ¼% of the measured value. For this experiment the Equivalent-Circuits Models yield
excellent estimates for the physical quantities.
Run Notes/Comments
The Thevin/Norton Equivalents are very accurate for this circuit. This
indicates that the BOTH the resistors and Volage-Source very closely
approximate the LINEAR Circuit-Component ideal forms
© Bruce Mayer, PE • Chabot College • 282219678 • Page 10