Help to write Kirchhoff's equations for a circuit

[amiras]

Homework Statement

Help me to write Kirchhoff's equations for this circuit. I'd like to think that I do understand this method, however Current sources giving me some problems.

Homework Equations

Currents:
Sum = 0For closed loop:
$\sum IR = \sum E$

The Attempt at a Solution

Since I am free to choose the currents directions, i choosed them in the picture with red arrows and named junctions and edges.

According to instructions, I need to write (junctions-1) = 7-1 = 6 equations for first Kirchhoff's law. And (all branches-(branches with current sources)-junctions-1) = 11-2-6=3 with second law. (is it really 3?, or some branches should be excluded)

Work:

I start with second law, because I am not sure how to use first one with current sources, I know that using second law, I can't pick contour with current sources.

So I take ABFGA: (I assume that current index is the same as rezistor)

R3*I3 + R5*I5 + I3*R3 = E3

Another contour BCEFB:

R6*I6 + R1*I1 - R5*I5 - R3*I3 = 0

And let's say BCDHB:

R6*I6 + I2*R2 - R3*I3 = E2

Is this is ok?

Please help to write equations for current.(Picture attached)

 

[gneill]

I would be tempted to turn J2 and R2, and J3 and R3 into their Thvenin equivalents and then write three simple KVL loop equations.

 

[tiny-tim]

hi amiras!

(try using the X₂ icon just above the Reply box )

So I take ABFGA: (I assume that current index is the same as rezistor)

R3*I3 + R5*I5 + I3*R3 = E3

Another contour BCEFB:

R6*I6 + R1*I1 - R5*I5 - R3*I3 = 0

And let's say BCDEB:

R6*I6 + I2*R2 - R3*I3 = E2

correct so far (except you keep writing "3" instead of "4")

(and yes, there's only 3 loops since the loops with a current source don't count)

Please help to write equations for current.

first you need to name an extra current between AB (or, easier, is to treat AB as a single point with 4 wires, instead of two points with 3 wires each)

similarly you need to name a current through E₂ and E₃

then you need a current (KCL) equation at every node (except of course where you're joining two nodes into one, as at AB)

 

[amiras]

I would be tempted to turn J2 and R2, and J3 and R3 into their Thvenin equivalents and then write three simple KVL loop equations.

I understand that, however I would like to try without it :)

hi amiras!

(try using the X₂ icon just above the Reply box )correct so far (except you keep writing "3" instead of "4")

(and yes, there's only 3 loops since the loops with a current source don't count)first you need to name an extra current between AB (or, easier, is to treat AB as a single point with 4 wires, instead of two points with 3 wires each)

similarly you need to name a current through E₂ and E₃

then you need a current (KCL) equation at every node (except of course where you're joining two nodes into one, as at AB)

Ok added currents Iab and Ie2 Ie3. Let's try without joining nodes, I now see that it would be very helpful. (question: we also can join nodes at D and M?)

And what could the 4th contour I am missing, could it be for example: HDEF?

So this is it, I assume positive is the currents that go in, and negative that goes out.

A: I3+J3-Iab=0
B: Iab-I4-I6=0
D: (Do i need to add current I_dm? Or I_dm=Ie2?)
M: (Same thing, I am not sure)
F: I1+I5-Ie3=0
K: Ie3-I3-J3=0
H: I4+Ie2-I5=0

 

[tiny-tim]

hi amiras!

(just got up :zzz: …)

question: we also can join nodes at D and M?

yes

And what could the 4th contour I am missing, could it be for example: HDEF?

you're not missing anything …

there are 4 loops, yes, but only three (any three) of them are independent …

if you add the KVL equations for three of them, it comes to minus the fourth, so the fourth one gives you no new information

the number of loops in a diagram (once you've removed the current-source loops) is the "obvious" number, the number of lines you have to cut to make it loopless ​

I assume positive is the currents that go in, and negative that goes out.

i'm not sure what you mean by that

positive is the same direction as the arrow, which as you said earlier you are free to choose

A: I3+J3-Iab=0
B: Iab-I4-I6=0
F: I1+I5-Ie3=0
K: Ie3-I3-J3=0
H: I4+Ie2-I5=0

yes, that's correct, but you also need one for node N

D: (Do i need to add current I_dm? Or I_dm=Ie2?)
M: (Same thing, I am not sure)

no, I_dm is not I_e2 …

you do need an extra I_dm (or preferably make D and M the same node)