Exercise 2: Millman s Theorem

Transcription

1 Exercise 2: Millman s Theorem EXERCISE OBJECTIVE When you have completed this exercise, you will be able to solve a circuit by applying Millman s theorem. You will verify your results by comparing calculated and measured data. DISCUSSION The superposition method is used to individually determine the effects of each source on a common circuit element and then combine them algebraically. Millman s theorem uses the sum of the branch currents and the sum of the conductances to help you determine the voltage across the branches. To use Millman s theorem, the circuit used in Exercise 1 is redrawn to show each of its branches. No electrical changes are made. R1 and its series voltage source form a branch. R2 and its series voltage source form a branch. R3 is placed across (in parallel with) the other branches to form the third branch. 166 Festo Didactic P0

2 The objective is to determine the common branch voltage, which is the voltage drop of R3. Once known, this voltage leads to a solution of all circuit currents and voltage drops. Millman s theorem uses branch currents and conductance (G) to solve the circuit. a. V S2 + V R2 b. V S1 + V R1 c. V R3 d. All of the above Festo Didactic P0 167

3 Are the branch voltages equal? a. yes b. no According to Millman s theorem, divide the sum of the parallel currents (branch currents) by the sum of the parallel conductances (1/R) to determine the voltage drop of R3. V R3 VR1 VR2 VR3 + + = R1 R2 R R1 R2 R3 Sum of the branch currents Sum of the branch conductants BRANCH 1 current is the voltage of V S1 divided by ohmic value of R1 (V S1 /R1); voltage over resistance yields current. 168 Festo Didactic P0

4 V S1 is negative with respect to circuit common. BRANCH 2 current equals V S2 /R2. BRANCH 3 current equals V S3 /R3. Because BRANCH 3 does not have a voltage source, its current is zero (0/R3 = 0). Festo Didactic P0 169

5 PROCEDURE Locate the SUPERPOSITION circuit block, and connect the circuit shown. Adjust each variable voltage source to 10.0 Vdc. The Millman equivalent circuit schematic is illustrated. V R3 equals the sum of the branch currents divided by the sum of the circuit conductances. 170 Festo Didactic P0

6 What is the total conductance of the circuit? G T = millisiemens (Recall Value 1) What is the value of R1 branch current? Use V S1 /R1. I R1 = ma (Recall Value 2) Based on the Millman equivalent circuit, what is the R3 branch current? I R3 = ma (Recall Value 3) Festo Didactic P0 171

7 What is the value of R2 branch current? Use V S2 /R2. I R2 = ma (Recall Value 4) Based on your calculated values and the given formula, what is the value of V R3? G T = ms (Step 3, Recall Value 1) I R1 = ma (Step 4, Recall Value 2) I R2 = ma (Step 6, Recall Value 4) V R3 = (I R1 + I R2 )/G T V R3 = Vdc (Recall Value 5) Based on the Millman solution for V R3, can you determine the actual circuit currents and voltage drops? a. yes b. no 172 Festo Didactic P0

8 Use your voltmeter to measure V R3. Are your results consistent with Millman s theorem? a. yes b. no Are your results consistent with the Millman solution? a. yes b. no CONCLUSION Use Millman s theorem to calculate the voltage across a common circuit element. Millman s theorem takes the form of current divided by conductance. Current divided by conductance (I/G), or multiplied by resistance (I x R), yields voltage. Festo Didactic P0 173

9 REVIEW QUESTIONS 1. Based on Millman s theorem, what is the sum of the branch currents? a ma b ma c. ±4.08 ma d. None of the above 2. Based on Millman s theorem, what is the sum of the branch conductances? a ms b ms c. ±5.74 ms d. All of the above 3. With respect to circuit common, what is the voltage drop across R3? a. 0.7 V b. 0.7 V c. Either of the above d. None of the above 4. The sum of the conductances a. increases as the source voltage is increased. b. does not change with changes in source voltage. c. decreases as the source voltage is decreased. d. equals zero when the source voltage is zero. 174 Festo Didactic P0

10 5. For the circuit shown, a. Millman s theorem cannot be applied because common is not in the proper place. b. V S1 is negative when applied to Millman s theorem. c. V S2 is positive when applied to Millman s theorem. d. Millman s theorem does not affect the polarity of V S1 or V S2 when they are measured with respect to circuit common. Festo Didactic P0 175