Test Your EQ (Engineering Quotient)

EQ #62

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 numerical solution for the equations?

To solve the equations directly, start by solving the third equation for Xc and substituting it into the other two equations:

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Xc = 1/2 Xa + 1/2 Xb + 1/2

21Xa – 10Xb – 5Xa – 5Xb – 5 = 1
-10Xa + 22Xb – 5Xa – 5Xb – 5 = -2

16Xa – 15Xb = 6
-15Xa + 17Xb = 3

Solve for Xa by multiplying the first equation by 17 and the second equation by 15 and then adding them:

272Xa – 255Xb = 102
-225Xa + 255Xb = 45

47Xa = 147 → Xa = 147/47

Solve for Xb by multiplying the first equation by 15 and the second equation by 16 and then adding them:

240Xa – 225Xb = 90
-240Xa + 272Xb = 48

47Xb = 138 → Xb = 138/47

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Finally, substitute those two results into the equation for Xc:

Xc = 147/94 + 138/94 + 47/94 = 332/94 = 166/47

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EQ #62

by Circuit Cellar Staff time to read: 1 min