How do I do this with KCL and KVL

[strk]

Homework Statement

http://i139.photobucket.com/albums/q292/strkthug/KCL/kclandkvl.png

Homework Equations

The Attempt at a Solution

I think since x2 is 2a and x3 is 3a then x5 should be 5a
and I have no idead where to go from there

 

[staticd]

First, do a mesh analysis. Then, do a node analysis. Just do it one step at a time and you'll get it.

 

[strk]

First, do a mesh analysis. Then, do a node analysis. Just do it one step at a time and you'll get it.

what is mesh analysis and node analysis? how do I even start that out

 

[staticd]

I assume that you've been taught what KCL and KVL are.

Therefore, you know that in each of the three loops (or, meshes), the sum of the voltages (i * r) or x, in this case equals zero. That give you three equations. Start there and see what you get.

Watch the signs!

 

[berkeman]

what is mesh analysis and node analysis? how do I even start that out

Since the problem asks you to use KCL and KVL, we will assume that you know what they are. strk was just using alternate common names for them:

KCL = node (nodal) analysis

KVL = mesh analysis

Please show us your work, if you'd like us to check your answers.

 

[Hosam_IT]

i tried to solve it by KVL using signs of currents and got these:
Vx3=-1
Ix5=-5
Vx2=-4
Ix4=-3
but i don't know wether its right or not

 

[staticd]

I don't have time to check right now. However, you should be able to verify your answers using KCL and/or Ohm's law. That is the normal method to show yourself that the value you have found is correct...

 

[staticd]

i tried to solve it by KVL using signs of currents and got these:
Vx3=-1
Ix5=-5
Vx2=-4
Ix4=-3
but i don't know wether its right or not

Okay, my answers are a little different.

For all of the currents that you don't know, you can do KCL --

So, we need to find Ix4 and Ix5.

For, Ix4, let's look at the node where x4,2, and 1 meet.

There you have...

1A + 2A = Ix4 --> Ix4 = 3A

for, Ix5, let's look at the node where x2, 3, and 5 meet.

There you have...

3A + 2A + Ix5 = 0 --> Ix5 = -5A

Now, you have three more voltages to find; Vx2, 3, and 1.

For the loop with x4, 2, and 5, you have...

-1V - 3V + Vx2 = 0 --> Vx2 = 4V

For the loop with x5 and x3, you have...

-Vx3 + 1 = 0 --> Vx3 = 1V

For the loop with x2, 3, and 1, you have...

Vx1 + 1V - 4V = 0 --> Vx1 = 3V

I hope that helps... Watch your signs! Start with what you KNOW and take small steps to find the solution!

 

[Hosam_IT]

Okay, my answers are a little different.

For all of the currents that you don't know, you can do KCL --

So, we need to find Ix4 and Ix5.

For, Ix4, let's look at the node where x4,2, and 1 meet.

There you have...

1A + 2A = Ix4 --> Ix4 = 3A

for, Ix5, let's look at the node where x2, 3, and 5 meet.

There you have...

3A + 2A + Ix5 = 0 --> Ix5 = -5A

Now, you have three more voltages to find; Vx2, 3, and 1.

For the loop with x4, 2, and 5, you have...

-1V - 3V + Vx2 = 0 --> Vx2 = 4V

For the loop with x5 and x3, you have...

-Vx3 + 1 = 0 --> Vx3 = 1V

For the loop with x2, 3, and 1, you have...

Vx1 + 1V - 4V = 0 --> Vx1 = 3V

I hope that helps... Watch your signs! Start with what you KNOW and take small steps to find the solution!

that's right you got same answers by using its directions

 

[staticd]

Right, Hosam. However, you have to know that 4V does not equal -4V! Very tricky. Very tricky, indeed.