Issue 282: EQ Answers

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

The circuit should be designed so that each ammeter displays one of the values Xa, Xb, or Xc. Given that the Xa, Xb, and Xc values represent currents, what kind of circuit analysis yields equations in this form?

ANSWER 1
You get equations in this form when you do mesh analysis of a circuit. Each equation represents the sum of the voltages around one loop in the mesh.

PROBLEM 2
What do the coefficients on the left side of the equations represent? What about the constants on the right side?

ANSWER 2
The coefficients on the left side of each equation represent resistances. Resistance multiplied by current (the unknown Xa, Xb, and Xc values) yields voltage.
The “bare” numbers on the right side of each equation represent voltages directly (i.e., independent voltage sources).

PROBLEM 3
What is the numerical solution for the equations?

ANSWER 3
To solve the equations directly, start by solving the third equation for Xc and substituting it into the other two equations:

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

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

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

PROBLEM 4
Finally, what is the actual circuit? Draw a diagram of the circuit and indicate the required value of each voltage source.

ANSWER 4
The circuit is a mesh comprising three loops, each with a voltage source. The common elements of the three loops are the three 10-Ω resistors, connected in a Y configuration (see the figure below).

cc281_eq_fig1The 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 ensure they’re self-consistent.

We’ll start by naming Va as ground, or 0 V:

Vb = Va + 2 V = 2 V

Vc = Vb + 2 Ω × Xb = 2V + 2 Ω × 138/47 A = 370/47 V = 7.87234 V

Vd = Vc + 1 Ω × Xa = 370/47 V + 1 Ω × 147/47A = 517/47 V = 11.000 V

Ve = Vd – 1 V = 11.000 V – 1.000 V = 10.000 V

Va = Ve – 10 V = 0 V

which is where we started.

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

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

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

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

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

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

Electrical Engineering Crossword (Issue 283)

The answers to Circuit Cellar’s February electronics engineering crossword puzzle are now available.

283-crossword-key

Across

2. LITZWIRE—Separately insulated strands woven together [two words]
4. LINKFIELD—First in a message buffer’s line [two words]
6. PETAFLOPS—Measures a processor’s floating point unit performance
8. ANION—negatively charged atom
9. LISP—Used to manipulate mathematical logic
11. STATCOULOMB—i.e., franklin (Fr)
12. AMBISONICS—Typically requires a soundfield microphone
15. BROUTER—This device can send data between networks and it can forward data to individual systems in a network
16. TRINITRON—This CRT technology was originally introduced the 1960s
17. OXIDE—The “O” in CMOS
18. DETENT—Used to prevent or stop something from spinning

Down

1. ELECTRICSUSCEPTIBILITY—XE [two words]
3. RADECHON—A barrier-grid storage tube
5. COLPITTS—This oscillator uses two-terminal electrical components to create a specific oscillation frequency
7. BEAGLEBOARD—TI’s open-source SBC
10. PICONET—A network that is created using a wireless Bluetooth connection
13. BETATRON—Designed to accelerate electrons
14. TEBIBYTE—More than 1,000,000,000,000 bytes

Peter Baston Wins the CC Code Challenge (Week 31)

We have a winner of last week’s CC Weekly Code Challenge, sponsored by IAR Systems! We posted a code snippet with an error and challenged the engineering community to find the mistake!

Congratulations to Peter Baston of Flintshire, United Kingdom for winning the CC Weekly Code Challenge for Week 31! Peter will receive a Circuit Cellar 2012 & 2011 Archive CD.

Peter’s correct answer was randomly selected from the pool of responses that correctly identified an error in the code. Peter answered:

Line 35: Should not end with semi-colon

2013_code_challenge_31_answer

You can see the complete list of weekly winners and code challenges here.

What is the CC Weekly Code Challenge?
Each week, Circuit Cellar’s technical editors purposely insert an error in a snippet of code. It could be a semantic error, a syntax error, a design error, a spelling error, or another bug the editors slip in. You are challenged to find the error. Once the submission deadline passes, Circuit Cellar will randomly select one winner from the group of respondents who submit the correct answer.

The CC Weekly Code Challenge ran from June 3rd through December 30th, 2013. Subscribe to our CC.Post newsletter to stay informed of other contests and challenges, as well as recent news, new issue availability, and more!

Gait Boxman Wins the CC Code Challenge (Week 30)

We have a winner of last week’s CC Weekly Code Challenge, sponsored by IAR Systems! We posted a code snippet with an error and challenged the engineering community to find the mistake!

Congratulations to Gait Boxman of Gelderland, Netherlands for winning the CC Weekly Code Challenge for Week 30! Gait will receive an IAR Kickstart: KSK-FM3-48PMC-USB.

Gait’s correct answer was randomly selected from the pool of responses that correctly identified an error in the code. Gait answered:

Line 31: should be digits % 3 instead of digits / 3

2013_code_challenge_30_answer

You can see the complete list of weekly winners and code challenges here.

What is the CC Weekly Code Challenge?
Each week, Circuit Cellar’s technical editors purposely insert an error in a snippet of code. It could be a semantic error, a syntax error, a design error, a spelling error, or another bug the editors slip in. You are challenged to find the error.Once the submission deadline passes, Circuit Cellar will randomly select one winner from the group of respondents who submit the correct answer.

Inspired? Want to try this week’s challenge? Get started!

Submission Deadline: The deadline for each week’s challenge is Sunday, 12 PM EST. Refer to the Rules, Terms & Conditions for information about eligibility and prizes.

Gordon Margulieux Wins the CC Code Challenge (Week 29)

We have a winner of last week’s CC Weekly Code Challenge, sponsored by IAR Systems! We posted a code snippet with an error and challenged the engineering community to find the mistake!

Congratulations to Gordon Margulieux of Oregon, United States for winning the CC Weekly Code Challenge for Week 29! Gordon will receive an Elektor 2012 & 2011 Archive DVD.

Gordon’s correct answer was randomly selected from the pool of responses that correctly identified an error in the code. Gordon answered:

Line 10: Conditional should be “if (number == 0)” instead of number < 0

2013_code_challenge_29_answer

You can see the complete list of weekly winners and code challenges here.

What is the CC Weekly Code Challenge?
Each week, Circuit Cellar’s technical editors purposely insert an error in a snippet of code. It could be a semantic error, a syntax error, a design error, a spelling error, or another bug the editors slip in. You are challenged to find the error.Once the submission deadline passes, Circuit Cellar will randomly select one winner from the group of respondents who submit the correct answer.

Inspired? Want to try this week’s challenge? Get started!

Submission Deadline: The deadline for each week’s challenge is Sunday, 12 PM EST. Refer to the Rules, Terms & Conditions for information about eligibility and prizes.