- #1

- 88

- 1

## Homework Statement

Please see a picture

## Homework Equations

KVL

## The Attempt at a Solution

V1-I1Z1-(I1-I2)Z4=0

Z2(I2-I4)+Z5(I2-I3)-(I2-I1)Z4=0

Z3I3+V2+Z5(I3-I2)=0

V3+Z2(I2-I4)=0

Any clues are appreciated..

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter M P
- Start date

- #1

- 88

- 1

Please see a picture

KVL

V1-I1Z1-(I1-I2)Z4=0

Z2(I2-I4)+Z5(I2-I3)-(I2-I1)Z4=0

Z3I3+V2+Z5(I3-I2)=0

V3+Z2(I2-I4)=0

Any clues are appreciated..

- #2

gneill

Mentor

- 20,945

- 2,886

When you write your mesh equations make sure to first indicate your assumed current directions, then make sure that you are consistent in summing either potential drops or potential rises. I note, for example, that last term in your second equation seems to be inconsistent. The first two terms look like potential drops due to assumed clockwise mesh currents, while the last term looks like a potential rise. Similarly for your last equation, the currents in the second term don't don't match the assumed mesh current directions if the V3 term is taken to be a potential drop for KVL going clockwise around the loop.## Homework Statement

Please see a picture

View attachment 76921

## Homework Equations

KVL

## The Attempt at a Solution

V1-I1Z1-(I1-I2)Z4=0

Z2(I2-I4)+Z5(I2-I3)-(I2-I1)Z4=0

Z3I3+V2+Z5(I3-I2)=0

V3+Z2(I2-I4)=0

Any clues are appreciated..

- #3

- 88

- 1

When you write your mesh equations make sure to first indicate your assumed current directions, then make sure that you are consistent in summing either potential drops or potential rises. I note, for example, that last term in your second equation seems to be inconsistent. The first two terms look like potential drops due to assumed clockwise mesh currents, while the last term looks like a potential rise. Similarly for your last equation, the currents in the second term don't don't match the assumed mesh current directions if the V3 term is taken to be a potential drop for KVL going clockwise around the loop.

Attached direction of currents as suggested.

- #4

gneill

Mentor

- 20,945

- 2,886

Usually it's convenient to have all the mesh currents be either clockwise or counterclockwise (there's a good reason for this, having to do with a very simple way to write the mesh equations in matrix form by inspection which I'm sure you'll cover in your course at some point).

You seem to have chosen the following:

I1 : clockwise

I2 : clockwise

I3 : counterclockwise

I4 : counterclockwise

That's fine, but now you'll need to write out your mesh equations accordingly.

- #5

- 88

- 1

Usually it's convenient to have all the mesh currents be either clockwise or counterclockwise (there's a good reason for this, having to do with a very simple way to write the mesh equations in matrix form by inspection which I'm sure you'll cover in your course at some point).

You seem to have chosen the following:

I1 : clockwise

I2 : clockwise

I3 : counterclockwise

I4 : counterclockwise

That's fine, but now you'll need to write out your mesh equations accordingly.

V1 -(Z1+Z4)J1+Z4J2 =0

Z4J1-(Z4+Z2+Z5)J2+Z5J3-V3=0

Z5I2-(Z5+Z3)I3-V2=0

- #6

- 88

- 1

V3+Z2(I4-I2)=0 ??

[

[

- #7

gneill

Mentor

- 20,945

- 2,886

Okay, so you've decided that Z2 is irrelevant to the second loop since its potential difference is constrained by V3. That's a good observation! It reduces the number of loops that you have to deal with. However, it means that Z2 should not appear in the second term of your second equation above. Make that change and try solving the system for the mesh currents.V1 -(Z1+Z4)J1+Z4J2 =0

Z4J1-(Z4+Z2+Z5)J2+Z5J3-V3=0

Z5I2-(Z5+Z3)I3-V2=0

- #8

- 88

- 1

Z4J1-(Z4+Z5)J2+Z5J3-V3=0

Z5I2-(Z5+Z3)I3-V2=0

V3+Z2(I4-I2)=0

Sorry I am a bit confused so I have now 4 equations as above with no Z2 in 2nd equation? does it make sense now??

- #9

gneill

Mentor

- 20,945

- 2,886

You don't need the fourth equation to find the desired ##I## . As I mentioned above, V3 fixed the potential drop across Z2, and I4 plays no role in determining ##I##.

Z4J1-(Z4+Z5)J2+Z5J3-V3=0

Z5I2-(Z5+Z3)I3-V2=0

V3+Z2(I4-I2)=0

Sorry I am a bit confused so I have now 4 equations as above with no Z2 in 2nd equation? does it make sense now??

If you want to look at it another way, recall that components in parallel share the same potential difference, and their order doesn't matter so long as they remain parallel-conected. So you can swap the locations of Z2 and V3 without altering the circuit behavior at all. That puts V3 in the second loop and removes Z2 from it. The current through Z2 is trivially given by Ohms law since you have the impedance and the potential difference, so no need to include loop ##I4## in your set of equations. The circuit then looks like this:

- #10

- 88

- 1

Words are powerless to express my appreciation :)

- #11

- 88

- 1

I1 = 0.613 - j26.4 A

I2 = 9.96 - j50.7 A

I3 = 26.8 - j20.3 A

- #12

- 88

- 1

sorry missed -0.613 -j26.4 for I1

- #13

gneill

Mentor

- 20,945

- 2,886

Hmm. Those don't look like the values that I'm seeing. What method are you using to solve the system of equations? I know it can be a tough slog to do the complex arithmetic by hand... If you look around online you can find "calculators" that do complex math, which you can use to confirm your steps.

I1 = 0.613 - j26.4 A

I2 = 9.96 - j50.7 A

I3 = 26.8 - j20.3 A

- #14

- 88

- 1

- #15

gneill

Mentor

- 20,945

- 2,886

I can't comment on what I cannot see. So you'll have to find a way to display your steps.

This calculator that you were given, is it a matrix-type of solver? If so, perhaps you could post your matrix equation (impedance and voltage matrices).

- #16

- 88

- 1

I3= 25.9 - j40.2

I2= 18.1 - j22.1

- #17

- #18

- 88

- 1

I cannot thank you enough for helping me:)

- #19

- #20

- 88

- 1

- #21

- 88

- 1

You're very welcome.

Let us know how the nodal analysis version of the problem goes.

3. The Attempt at a Solution

node 1 = V10 etc.

V10-V30/Z1 + V40-V30/Z2 + 0-V30/Z4 = 0

V20-V40/Z3 + V30-V40/Z2 + 0-V40/Z5 = 0

thank you in advance

- #22

gneill

Mentor

- 20,945

- 2,886

I'm not seeing how your equations pertain to the circuit. Perhaps it would help if you were to indicate on your circuit diagram the nodes. Remember that you need to identify a reference node ("ground") and the essential nodes. Hint: V3 makes life easier once again in a super way.## The Attempt at a Solution

node 1 = V10 etc.

V10-V30/Z1 + V40-V30/Z2 + 0-V30/Z4 = 0

V20-V40/Z3 + V30-V40/Z2 + 0-V40/Z5 = 0

thank you in advance

- #23

- 88

- 1

- #24

gneill

Mentor

- 20,945

- 2,886

That leaves nodes 3 and 4, both of which are where several branches supporting different currents connect. This identifies them as essential nodes. You did not identify your reference node, but I am going to assume that you've selected the bottom node for that purpose, correct?

Now, go back to the hint I left you with last time. What can you say about the relationship between the potentials at nodes 3 and 4?

- #25

- 88

- 1

another attempt:

V1= (V1-0)/Z4 + (V2-0)/Z5 +V2

-V1-V3+V2=0

V1= (V1-0)/Z4 + (V2-0)/Z5 +V2

-V1-V3+V2=0

- #26

gneill

Mentor

- 20,945

- 2,886

For nodal analysis you want to write KCL at each node. That means summing currents either entering or leaving a given node. You use the node voltages to determine the potential across a given branch, and then use that branch's impedance with Ohm's Law to reckon the current.

Nodes connected to each other by a voltage source are a special case (hence the hint).

- #27

- 88

- 1

- #28

gneill

Mentor

- 20,945

- 2,886

Getting closer. However...V30-V1/Z1 + V30-V40/Z2+V30/Z4+V40-V30/Z2+V40-V2/Z3+V40/Z5 =0

V40=V30+V3

V3 forces a fixed potential difference between V30 and V40. This creates a special situation. Note that the voltage source V3 has no impedance associated with it, so you cannot treat it like another branch and write a current for it via Ohm's Law. Yet it connects V30 and V40, and must influence KCL at those nodes.

This is a case where you consider that the fixed relationship between those nodes imposed by that fixed voltage source creates what is called a

Check your text, notes, or web for the supernode concept. See if you can't identify the supernode on your circuit diagram and draw a boundary around it. When you write the node equation for a supernode you consider only the currents entering or leaving the supernode as a whole (passing through the boundary, as it were). Currents internal to the node are ignored, and only the fixed potential differences have an impact (setting the potentials at the wires protruding from the supernode).

- #29

- 88

- 1

Lets stick to web. Let me just ask a quick question. my equation need to be equal to V30-V40/Z2?

- #30

- 88

- 1

Getting closer. However...

V3 forces a fixed potential difference between V30 and V40. This creates a special situation. Note that the voltage source V3 has no impedance associated with it, so you cannot treat it like another branch and write a current for it via Ohm's Law. Yet it connects V30 and V40, and must influence KCL at those nodes.

This is a case where you consider that the fixed relationship between those nodes imposed by that fixed voltage source creates what is called asupernode. A supernode is made up of a set of nodes with potential differences that are fixed by voltage sources connected between them.

Check your text, notes, or web for the supernode concept. See if you can't identify the supernode on your circuit diagram and draw a boundary around it. When you write the node equation for a supernode you consider only the currents entering or leaving the supernode as a whole (passing through the boundary, as it were). Currents internal to the node are ignored, and only the fixed potential differences have an impact (setting the potentials at the wires protruding from the supernode).

V30-V1/Z1+V30/Z4+V40/Z5+V40-V2/Z3 = 0

V30-V40=V3

that is what I understood

Share:

- Replies
- 15

- Views
- 2K