Test Your EQ (Engineering Quotient)

EQ #63

Construct an electrical circuit to find the values of Xa, Xb, and Xc in this system of equations:

21Xa – 10Xb – 10Xc = 1
–10Xa + 22Xb – 10Xc = –2
–10Xa – 10Xb + 20Xc = 10

Your circuit should include only the following elements:

  • one 1Ω resistor
  • one 2Ω resistor
  • three 10Ω resistors
  • three ideal constant voltage sources
  • three ideal ammeters

The circuit should be designed such that each ammeter displays one of the values, Xa, Xb and Xc.

What is the actual circuit? Draw a diagram of the circuit and indicate the required value of each voltage source.

The circuit is a mesh comprising three loops, each with a voltage source. The common elements of the three loops are the three 10Ω resistors, connected in a Y configuration.

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The values of the voltage sources in each loop are given directly by the equations, as shown.

To verify the numeric solution calculated previously, you can calculate all of the node voltages around the outer loop, plus the voltage at the center of the Y, and make sure they’re self-consistent.

We’ll start by naming Va as ground, or 0V.
Vb = Va + 2V = 2V
Vc = Vb + 2Ω*Xb = 2V + 2Ω*138/47A = 370/47V = 7.87234V
Vd = Vc + 1Ω*Xa = 370/47V + 1Ω*147/47A = 517/47V = 11.000V
Ve = Vd – 1V = 11.000V – 1.000V = 10.000V
Va = Ve – 10V = 0V, which is where we started.

The center node, Vf, should be at the average of the three voltages Va, Vc and Ve:

0V + 370/47V + 10V / 3 = 840/141V = 5.95745V

We should also be able to get this value by calculating the voltage drops across each of the three 10Ω resistors:

Va + (Xc – Xb)*10Ω = 0V + (166-138)/47A * 10Ω = 280/47 V = 5.95745V

Vc + (Xb – Xa)*10Ω = 370/47V + (138-147)/47A * 10Ω = 280/47 V = 5.95745V

Ve + (Xa – Xc)*10Ω = 10V + (147-166)/47A * 10Ω = 280/47 V = 5.95745V

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EQ #63

by Circuit Cellar Staff time to read: 1 min