Electronic – In mesh analysis how are currents negative and positive

circuit analysiskirchhoffs-laws

I'm trying to do mesh analysis for the following circuit. I take the clockwise direction for the current as it's general convention.

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The reference shows that this results in

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I don't understand how the +,- terminals are marked for resistance and how VS1 becomes negative. Can someone help?

Best Answer

@elecman, let's work this out together and see how they got what they got.

The +ve and -ve terminals for Vs1 and Vs2 are noted by their longer and shorter lines. Longer line is for +ve terminal, shorter line is for -ve terminal.

Now when doing a Mesh Analysis, (IF NOT PREVIOUSLY STATED) the polarity of the resistor depends on the direction of the current going through that resistor.

Basically there is a voltage drop from the point when the current enters the resistor and the point when it leaves the resistor. Resistors "resist (more like reduce)" current flow, and thus lower the voltage levels within a circuit.

So for Loop 1, we have the current I1. Since I1 enters the Resistor R1 on the left side of the zigzag first, the left side is considered +ve and since it leaves the right side of the zigzag last, the right side of Resistor R1 is considered -ve. So If I1 was going in the opposite direction, thus entering the resistor R1 on the right side of the zigzag and leaving at the left side of the zigzag, the polarities of R1 will be opposite.

As for Resistor R2 we have a current I3. I3 however depends on the final battle/result between currents I1 and I2. Now I3 can go either way, it can enter R2 from the top and leave at the bottom or it can enter from the bottom and leave at the top. Since we are still in Loop 1, we want I3 to go in the same direction of I1 to keep everything simple, thus we want it to enter the top of Resistor R2 (making the top of the zigzag the +ve) and leave at the bottom of R2 (making the bottom of the zigzag the -ve). However in order for I3 to do that, I1 must be stronger than the opposite current going through that same resistor which is I2. Thus as for now in Loop 1, I3 = (I1 - I2).

Now we want to come up with the mesh equation for Loop 1: Following in the direction of current I1,

-Vs1 + I1 * R1 + (I1 - I2) * R2 = 0

Vs1 = I1 * R1 + (I1 - I2) * R2 --------- EQUATION 1

As for Loop 2, we have the current I2. Since we want the current entering R2 to be in the direction of I2 (from bottom to top making the bottom terminal of the zigzag of Resistor R2 +ve and the top terminal of the zigzag +ve), current I2 must be stronger than the opposite current I1. Thus in this case I3 = (I2 - I1).

As for Resistor R3, the current I2 is entering the left side of the zigzag, thus making that side the +ve terminal and leaving the right side of the zigzag, thus making that side the -ve terminal.

Now we want to come up with the mesh equation for Loop 2: Following in the direction of current I2,

(I2 - I1) * R2 + I2 * R3 + Vs2 = 0

-Vs2 = I2 * R3 + (I2 - I1) * R2 --------- EQUATION 2

I GUESS THAT R1 IN THE SECOND EQUATION FROM THEIR RESULTS SHOULD BE R3.