# KVL Mesh Current Method

#### Steve1992

Joined Apr 7, 2006
100
Using the KVL Mesh Current Method from the AAC e-textbooks.

The loop direction should tell you which way the current is flowing.
ie positive current: right direction. Negative current means re-draw arrow.

But I get a positive current from either direction for the loops:

Loop1
12V - R2I1 - 3.3V = 0
I1 = 17 mA

Loop1 (drawn other way)
-12V + 3.3V + R2I1 = 0
I1 = 17 mA

Where am I going wrong?

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#### steveb

Joined Jul 3, 2008
2,431
Using the KVL Mesh Current Method from the AAC e-textbooks.

The loop direction should tell you which way the current is flowing.
ie positive current: right direction. Negative current means re-draw arrow.

But I get a positive current from either direction for the loops:

Loop1
12V - R2I1 - 3.3V = 0
I1 = 17 mA

Loop1 (drawn other way)
-12V + 3.3V + R2I1 = 0
I1 = 17 mA

Where am I going wrong?
When you change the direction of the current, you need to change the order of the + and - terminals on the resistor. In other words, use the same current direction convention in both cases. In the first calculation you did, you used the positive current convention, with the arrow going from + to -. In the second example, you used the negative current convention, with the arrow going from - to +.

#### Ratch

Joined Mar 20, 2007
1,068
Using the KVL Mesh Current Method from the AAC e-textbooks.

The loop direction should tell you which way the current is flowing.
ie positive current: right direction. Negative current means re-draw arrow.

But I get a positive current from either direction for the loops:
Certainly you do. You should. You applied Kirchoff's voltage law (KVL) which is correct, and gives the same result no matter which direction you transit the loop. Notice that both equations are the same except for sign changes on both sides. You received a positive value for the mathematical result, which confirms your assumption that the positive current is existing in a clockwise direction. If you want to determine the real physical direction of the charge carriers, the negative electrons going counter-clockwise are the same mathematically as positrons going clockwise. So the real charge carriers (electrons) are going counter-clockwise.

Ratch

#### Steve1992

Joined Apr 7, 2006
100
Hi,

Im still struggling. Ive included a similar example.
By using the Mesh/loop method I correctly calculated currents:
I1 = 17 mA.
I2 = 5 mA.

My SPICE simulator says I3 is 22 mA. I know that would be the sum of I1 and I2, but I dont know how to get there?

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#### Ratch

Joined Mar 20, 2007
1,068
Steve1992,

My SPICE simulator says I3 is 22 mA. I know that would be the sum of I1 and I2, but I dont know how to get there?
OK, lets set up the equations for the left loop. Assuming a clockwise direction for the current and taking KVL counterclockwise, we get

-3.3 + 12 + 500*I1 = 0 =======> -17.4 ma . The minus sign means that the assumed direction for I1 was wrong, and the calculated current direction is counterclockwise.

For the right loop, we get -12 + 2200*I2 = 0 =======> I2 = 5.45 ma with the current direction agreeing with the assumption of clockwise movement. So both currents from each loop are going north through the 12 volt source. Therefore the sum of I1 and I2 equals I3 = 17 ma going north.

Ratch

#### hgmjr

Joined Jan 28, 2005
9,029
I was thinking that the current in right loop is governed only by the 3.3V voltage source divided by the 2200 ohm resistor. That would make the current in the right loop 1.5 millAmps.

hgmjr

#### bull1894

Joined Oct 18, 2009
4
the answer is 22.85 mA. the 12 volt battery is the greater source. the 3.3 volt battery works as a voltage resister. droping the 12 volts across it to 8.7 volts so that the 500 ohm resister has 17.4 mA at 8.7 V and standard 5.45 mA at 12V at the 2.2K resister.

#### hgmjr

Joined Jan 28, 2005
9,029
I was looking at how the problem would shape up if one applies super-position analysis. If you replace the 3.3V source by a short circuit, the current due to the 12V source does not produce any voltage drop across the 2200 resistor. Then if you short the 12V source to determine the impact of the 3.3V source you get the 1.5 millamps in the 2200 resistor due to the 3.3V source.

hgmjr

#### bull1894

Joined Oct 18, 2009
4
hgmjr

whats good for the goose is good for the gander, so the product of both shorts is the same.

break the curcit down to the sub parts and analize the interaction back to the source.

#### hgmjr

Joined Jan 28, 2005
9,029
hgmjr

whats good for the goose is good for the gander, so the product of both shorts is the same.

break the curcit down to the sub parts and analize the interaction back to the source.
I'm not sure what you mean.

hgmjr

#### Steve1992

Joined Apr 7, 2006
100
Thanks all.

Ive found using the Branch Current Method is the best solution.

#### hgmjr

Joined Jan 28, 2005
9,029
Thanks all.

Ive found using the Branch Current Method is the best solution.
Out of curiosity, I would be interested to learn what answer you got for the current in the left loop and the right loop?

hgmjr

#### Steve1992

Joined Apr 7, 2006
100
Out of curiosity, I would be interested to learn what answer you got for the current in the left loop and the right loop?

hgmjr
Left loop (I1) 17 mA
Right loop (I2) 5 mA
I3 = 22 mA

#### hgmjr

Joined Jan 28, 2005
9,029
To check my answer, I simulated the circuit in using Multisim. It came up with 17.4 milliamps in the left loop and 1.5 milliamps in the right loop. This is consistent with my earlier reply that the 2200 ohm resistor is experiencing a current of 1.5 milliamps.

hgmjr

#### Ratch

Joined Mar 20, 2007
1,068
hgmjr,

To check my answer, I simulated the circuit in using Multisim. It came up with 17.4 milliamps in the left loop and 1.5 milliamps in the right loop. This is consistent with my earlier reply that the 2200 ohm resistor is experiencing a current of 1.5 milliamps.
I agree with you, hgmjr. I mistakenly used 12 volts instead of 3.3 like I should have. So 1.5 ma plus 17.4 ma makes 18.9 ma of I3 current.

Ratch

#### hgmjr

Joined Jan 28, 2005
9,029
hgmjr,

I agree with you, hgmjr. I mistakenly used 12 volts instead of 3.3 like I should have. So 1.5 ma plus 17.4 ma makes 18.9 ma of I3 current.

Ratch
That is easily done.

I am not clear on the definition of the loop associated with your current I3. There are only two loops as far as I can tell so I would only expect to need to calculate two currents. The right loop stands pretty much on its on since the 3.3V ideal voltage source does not allow any of the current from the left loop to contribute in any way to the right loop's current.

hgmjr

#### steveb

Joined Jul 3, 2008
2,431
That is easily done.

I am not clear on the definition of the loop associated with your current I3.
The current I3 is the current in the 3.3V power supply. It is supposed to be found be adding the loop currents, since both loop currents flow through this common path.

The thing is that the math is still wrong. The mesh currents are opposing and the net current in the 3.3 V supply is -16 mA. The negative sign means that it is dissipating power and not supplying power. All power comes from the 12V supply in this example.

I tried to point out the importance of getting the mesh current direction correct in post number 2. The OP original question was that he tried the left mesh current in both counter-clockwise and clockwise directions but got a positive current in both cases, which is wrong.

This problem is quite trivial since the voltage on both resistors is seen by inspection. So the answer is obvious, but the purpose of the problem should be to get the correct answer by properly applying mesh current analysis.

I think some of the confusion is coming in since two schematics were shown with the power supplies swapped

#### JoeJester

Joined Apr 26, 2005
4,074
We have a difference of opinion between two simulators.....

Tina has I1 as 17.4 mA, I2 as 5.45 mA, and I3 as 22.85 mA, but I'm only using two decimal places on the meters.

Superposition has I1 as 17.4 mA, I2 as 5.455 mA and I3 as 22.855 mA

On edit ... maybe not since I used the "first" diagram and not the second. I'll check out the second one.

Tina has I1 as 17.4 mA, I2 as neg 1.5 mA, and I3 as 15.9 mA.

Superposition agrees with Tina ...

Last edited:

#### hgmjr

Joined Jan 28, 2005
9,029
Joe,

If you consider that the 3.3V ideal voltage source is the only voltage that can be seen by the 2200 ohm resistor, how can the current in the right loop be anything but 1.5 milliamps?

hgmjr

#### steveb

Joined Jul 3, 2008
2,431
We have a difference of opinion between two simulators.....

Tina has I1 as 17.4 mA, I2 as 5.45 mA, and I3 as 22.85 mA, but I'm only using two decimal places on the meters.

Superposition has I1 as 17.4 mA, I2 as 5.455 mA and I3 as 22.855 mA
Do we really need simulators for such a simple problem

I1=(12V-3.3V)/0.5kOhms=17.4 mA

I2=(12V)/2.2kOhms=5.454545... mA

I3=22.854545.... mA

the 45's are repeating numbers in the decimal.