# Source transformation on circuit with pairs of parallel current sources and resistors

• Engineering

## Homework Statement

The problem is attached below.

## The Attempt at a Solution

I tried reducing the circuit by applying a source transformation to each pair of current sources and resistors. I reduced the CS and R at the upper left to an 8V source and 4-ohm resistor in series, the CS and R at the upper right to a -4V source and 2 ohm-resistor in series, the CS and R in the bottom left to a -10V source and 4-ohm resistor in series, and the CS and R in the bottom right to a 10V source and 2-ohm resistor in series. After all these source transformations, I'm left with a circuit with 4 voltage sources and four resistors all in series, which I know how to solve. I think I might be applying these source transformations incorrectly, because circuitlab (online circuit simulator) is giving me a different answer than the one I have.

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NascentOxygen
Staff Emeritus

I tried reducing the circuit by applying a source transformation to each pair of current sources and resistors. I reduced the CS and R at the upper left to an 8V source and 4-ohm resistor in series, the CS and R at the upper right to a -4V source and 2 ohm-resistor in series, the CS and R in the bottom left to a -10V source and 4-ohm resistor in series, and the CS and R in the bottom right to a 10V source and 2-ohm resistor in series. After all these source transformations, I'm left with a circuit with 4 voltage sources and four resistors all in series, which I know how to solve.
Hi uasaki! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

So, what answer did you get for the loop current? Then, once you know the current into the terminals of your current-source-with-parallel-resistor, you can calculate how much current must be passing through that resistor.

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gneill
Mentor

When you transform a source and resistor, the "new" resistor does not function in the same way as in the original circuit --- it will carry a different current and show a different voltage drop.

So the first step here would be to transform all the other supplies and resistances first (you don't care about their particulars), then look at the new arrangement for additional simplifications.

Edit: The other option here is, when you make the transformation, to keep track of the nodes where Va is measured. When a current/resistor circuit is transformed to its voltage/resistor counterpart, the nodes in question end up spanning the new voltage source and resistor. Calculate Va accordingly, taking into account both the source voltage and the potential change across the resistor.

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Thanks for the help guys. I originally thought that the voltage drop across the resistor in the CS + R configuration was equal to the voltage drop across the resistor in the VS + R configuration. I didn't realize that the voltage drop and current across the resistor changed when you performed a source transformation.

Ok, does the polarity of the voltage drop across the resistor change how we do source transformation and/or apply the voltage divider rule in any way?

NascentOxygen
Staff Emeritus

Ok, does the polarity of the voltage drop across the resistor change how we do source transformation and/or apply the voltage divider rule in any way?
When you replace a CS+R with VS+R the polarity at its terminals does not change (and it follows that the direction of the current through the resistor does not change). I don't think you'll use voltage divider to solve this question.

There's a question back here are you able to answer it now?

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I'm not really sure how to find the loop current. Would writing KCLs at each of the nodes and/or a KVL around the resistors help me to find the loop current? Thanks for the help.

NascentOxygen
Staff Emeritus

You indicated that you know how to replace each CS+R with a VS+R. So do that for each individually, and clearly mark its two terminals so you are clear about what you've done. Then for all 4 sources connect those terminals up to form the circuit that you started with. What does it look like now?

It looks like a circuit with four voltage sources and four resistors in series. I can simplify this circuit and use the voltage divider rule to find Va but I don't think my polarities are correct. For my final circuit, I have a 8V source (- at the top, + at the bottom) in series with a 4-ohm resistor and an 8-ohm resistor. I'm also not sure why I would need to find the loop current in this case.

NascentOxygen
Staff Emeritus

It looks like a circuit with four voltage sources and four resistors in series.
Bingo! What is the total loop voltage? What is the total loop resistance?

The current through the circuit would just be the total voltage / total resistance. Then, I can just multiply the loop current by the value of the resistor. The problem is that the answer that I get from doing that is different than the answer that circuitlab is giving me. Also, wouldn't the current through the 4-ohm resistor (in the simplified circuit) be different from the current through the 4-ohm resistor in the original circuit?

NascentOxygen
Staff Emeritus

The current through the circuit would just be the total voltage / total resistance.
You're right to this point. What current do you calculate?
Then, I can just multiply the loop current by the value of the resistor.
No, you don't do that. You then go back to the original CS+R and say "With this value of current into its terminals, how much current goes through R?" After all, that's what you are trying to answer. But isn't the loop current equivalent to the current that flows through the 4-ohm resistor?

Edit: Nevermind, I think I got it. The current through the resistor in the original CS + R configuration would just be 2A - (-0.6) = 2.6 A.

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NascentOxygen
Staff Emeritus
You could now try working it without making the CS⟶VS conversion, as a check. You should get the same answer. 