Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

does my analisis of the circuit correct?

Status
Not open for further replies.
here is a solution with different resolt then mine
i cant understand where is my mistake
https://i50.tinypic.com/1rul4p.jpg

we need to find V1 and V2 and i_phi
i signed the potentials on the intersection as v1 and v2 .
kcl for v1 intersection:
[latex]
-\left ( \frac{8i_{\phi}-v_1}{2} \right )+\frac{v_1}{10}-\left ( \frac{v_2-v_1}{5} \right )=0
[/latex]
[latex]
-\left ( \frac{8\left ( \frac{v_2-v_1}{5} \right )-v_1}{2} \right )+\frac{v_1}{10}-\left ( \frac{v_2-v_1}{5} \right )=0\\
[/latex]

kcl for v2 intersection:
[latex]
-\left ( \frac{20-v_2}{2} \right )+\frac{v_2}{20}-\left ( \frac{v_2-v_1}{5} \right )=0\\
[/latex]
i got v1=20
v2=24
i_phi=4/5

my currents are pointed exactly as in the solution except the current which comes from the voltage source
in my solution it goes in the intersection.

i say that a current which goes inside signed as + and
its signed as - otherwise.

and as current goes the potential decreases
so i subtracted to higher potential with the lower potential divided by the ressistor between the
and put a minus if the current goes inside the intersection
 
Last edited:
I’m afraid that you do not have it quite right. You have 3 unknowns V1, V2 and Iphi. Therefore, you have to have 3 independent equations. I assume that the diamond on the right represents a constant current source that is there just to confuse the issue.

The other question you pose is the other analysis is using electron flow instead of current flow so the polarity will be the reverse of normal engineering results.
 
on the network v1 & v2 maked by red colour is not correct, the solution given on the link has considered v2 as v1 and v1 as v2, both are reversed.

any how you should come to the same solution, but i feel you bit confused about direction of current etc.

just stick on this "sum of current comming towords a node is zero"

also 8 i_phi is confusing? is it a voltage? then how it came in the picture? how it exactly relates to the i_phi? if its a part of current controlled voltage source then it can be correct.

just work it out again you will get the correct answers.

just check your second equation ....+v2/20 - (...) this should be v2/20 + (..)
 
Last edited:
Yes, Gary B is totally wrong; Iphi is defined as (V2-V1)/5 so that is the third equation. Your solution is correct, just check it again, buy again I think you have the right answers.
 
I assume that the diamond on the right represents a constant current source that is there just to confuse the issue

Not quite. The diamond represents a controlled or dependant voltage source with the control equation 8phi. This means the voltage accross it will be directly proportional to the current phi.
 
Okay, well unfortunately, without a sample problem or some type of question about those two techniques, it would be kinda hard to help. ;)
 
suggestion of book

please concern introductory circuit analysis by BOYLESTED(10th edition).
hope this will help u to clear your concepts.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top