Simple circuit analysis - can someone check if it's right?

In summary, the conversation discusses a simple circuit analysis and a solution attempt involving writing equations, solving for voltages, and determining currents using KVL and KCL. There is confusion about the correct approach and calculations, with the suggestion to assume V1, V2, and V3 are equal and use Ohm's Law and KCL to solve. There is also mention of the windows separation method, but it is not clear how to apply it.
  • #1
ABoul
28
0
simple circuit analysis -- can someone check if it's right?

Homework Statement



(circuit can be found on http://www.MailFreeOnline.com/uploader/A723E91C.jpg )

i. write down equations relating i1 and v1, i2 and v2 and i3 and v3
ii. solve for the voltages v1, v2 and v3
iii. determine the currents i1, i2 and i3


Homework Equations


KVL
KCL


The Attempt at a Solution


i.

i1 = 3 - V2
i2 = 2 - 2V2
i3 = V3/10


ii.
i1 + i2 = i3
3 - 2V1 + 2 - 2V2 = V3/10
therefore 5 - 2(V1 + V2) = V3/10

iii.
(this is the hard part for me. have i used KVL right?)
first window:
-1.5 + 1 + V1 + V2 = 0
therefore v1 + v2 = 0.5

second window:
-1 + V2 + v3 = 0
therefore V2 + v3 = 1

overall window:
V1 + V2 + V3 = 1.5
 

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  • #2


hey if you host your image on imageshack.us and past the link into the forum, the image can be viewed in the post without the approval of the admin. not saying that the admin will block it, but it will allow the image to be viewed sooner

if i see the image i may be able to help
 
  • #3


pavadrin said:
hey if you host your image on imageshack.us and past the link into the forum, the image can be viewed in the post without the approval of the admin. not saying that the admin will block it, but it will allow the image to be viewed sooner

if i see the image i may be able to help

hey. sorry about that. here's a link to the circuit: http://www.MailFreeOnline.com/uploader/A723E91C.jpg
 
Last edited by a moderator:
  • #4


ABoul said:
i1 = 3 - V2
i2 = 2 - 2V2
i3 = V3/10

Think about what you did for I1 and I2, the left side of the equation is in the units of Ampere, and the right side of the equation is in the units of Voltage.

Stick with Ohm's Law to find I1 and I2, similar to what you did for I3
 
  • #5


noumed said:
Think about what you did for I1 and I2, the left side of the equation is in the units of Ampere, and the right side of the equation is in the units of Voltage.

Stick with Ohm's Law to find I1 and I2, similar to what you did for I3

huh? isn't i1 = (1.5 - V1)/0.5, for example?
 
  • #6


You're right, sorry about that.

So how did you get I1 = 3 - V2? Solving it your way, wouldn't it be 3 - 2V1?
 
  • #7


noumed said:
You're right, sorry about that.

So how did you get I1 = 3 - V2? Solving it your way, wouldn't it be 3 - 2V1?

whoops sorry... it must have been a typo. how about part 3?
 
  • #8


Looking at your steps, it looks like you're doing okay. But somehow I can't agree with you that V1 + V2 + V3 = 1.5. I don't think you can combine it like that. I want to say that V1 = V2 = V3 because they're the same node.

(but then your equation V1 + V2 = 0.5 and V2 + V3 = 1 no longer makes sense...)
 
  • #9


noumed said:
Looking at your steps, it looks like you're doing okay. But somehow I can't agree with you that V1 + V2 + V3 = 1.5. I don't think you can combine it like that. I want to say that V1 = V2 = V3 because they're the same node.

(but then your equation V1 + V2 = 0.5 and V2 + V3 = 1 no longer makes sense...)

hmm... what does that mean?
 
  • #10


Perhaps redrawing the circuit will help you.
 

Attachments

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  • #11


noumed said:
Perhaps redrawing the circuit will help you.

ah i see what you mean now. so where did i go wrong? i thought my KVL logic was correct...
 
  • #12


also, isn't it true that i1 + i2 = i3? if you plug in my values, you get something else.
 
  • #13


I vaguely remember the windows separation method, like what you did. But I think you're supposed to follow the path of the current, so you can't do something like

-1.5 + V1 + V2 + 1 = 0

Instead, you're supposed to go to R3,

-1.5 + V1 + V3 = 0

It's been a while since I've done it, so maybe? heheh.. but if you look at the circuit I came up with, it should be more clear.

And yes, I1 + I2 = I3. What do you mean by that?
 
  • #14


noumed said:
I vaguely remember the windows separation method, like what you did. But I think you're supposed to follow the path of the current, so you can't do something like

-1.5 + V1 + V2 + 1 = 0

Instead, you're supposed to go to R3,

-1.5 + V1 + V3 = 0

It's been a while since I've done it, so maybe? heheh.. but if you look at the circuit I came up with, it should be more clear.

And yes, I1 + I2 = I3. What do you mean by that?

if you substitute in my (supposedly correct) equations for i1, i2 and i3, you don't reach the conclusion that V1 = V2 = V3...

also, using the window method, I'm sure you can inspect every window individuall. i just don't know how to do it lol.
 
  • #15


I do believe the right way to solve this is to assume that V1 = V2 = V3. Let's look at the windows of the final circuit I drew then.

I1 = (Vs1 - V3) / R1
I2 = (Vs2 - V3) / R2

But we also know that V3 = I3 * R3 by Ohm's Law.

By KCL, we know that I3 = I1 + I2.

V3 = (I1 + I2) * R3 and plug in I1 and I2 from above, and you should have V3, which is equals to V1 and V2. From there you should be able to solve for I1 and I2.
 
  • #16


noumed said:
I do believe the right way to solve this is to assume that V1 = V2 = V3. Let's look at the windows of the final circuit I drew then.

I1 = (Vs1 - V3) / R1
I2 = (Vs2 - V3) / R2

But we also know that V3 = I3 * R3 by Ohm's Law.

By KCL, we know that I3 = I1 + I2.

V3 = (I1 + I2) * R3 and plug in I1 and I2 from above, and you should have V3, which is equals to V1 and V2. From there you should be able to solve for I1 and I2.

ah thansk for that. I'm just waiting to see your circuit (it's still being approved).
 
  • #17
ABoul said:
ah thansk for that. I'm just waiting to see your circuit (it's still being approved).

d'oh! didn't read the rules =]

http://rapidshare.com/files/159049612/ckt.pdf.html

for your convenience..
 
  • #18


Could anyone solve this problem
 

Attachments

  • Circuit.JPG
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Related to Simple circuit analysis - can someone check if it's right?

1. How do you analyze a simple circuit?

To analyze a simple circuit, you first need to identify all the components in the circuit, such as resistors, capacitors, and voltage sources. Then, you can use Ohm's law and Kirchhoff's laws to calculate the voltage, current, and resistance at different points in the circuit. Finally, you can use these values to determine the overall behavior of the circuit.

2. What is the purpose of circuit analysis?

The purpose of circuit analysis is to understand how a circuit works and how its components affect the flow of electricity. By analyzing a circuit, we can determine the voltage, current, and power at different points in the circuit, and make predictions about its behavior.

3. Can I use circuit analysis for more complex circuits?

Yes, circuit analysis can be used for more complex circuits as well. The principles of Ohm's law and Kirchhoff's laws apply to all circuits, regardless of their complexity. However, the calculations may become more tedious and time-consuming for larger and more complex circuits.

4. What are the common mistakes to avoid in circuit analysis?

Some common mistakes to avoid in circuit analysis include forgetting to account for the direction of current flow, not taking into consideration the internal resistance of components, and not considering the effects of non-ideal components. It is also important to double-check your calculations and make sure they make sense in the context of the circuit.

5. Can someone check if my circuit analysis is correct?

Yes, it is always a good idea to have someone else check your circuit analysis to ensure accuracy. This can be a fellow scientist, a professor, or even an online community. Getting a second opinion can help catch any mistakes or oversights and improve your understanding of circuit analysis.

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