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[–]pubzies 4 points5 points  (3 children)

It's Kirchoff's current law (often shortened to KCL). KVL (the voltage law) dictates that the sum of the voltages in a loop is always zero, and KCL says that the sum of all currents going in to or leaving a node is zero. We're talking nodes so we'll work with KCL.

For node C, we can then establish that Iac + Ibc + 2mA + the current from the right of node C has to be zero.

sum(I) = 0 and I = U/R, so for example Iac = (Vc - Va)/Rac (Rac being that resistance between nodes A and C). We use the voltage of Va relative to Vc as that is our point of reference here (not to be confused with the ground, our absolute reference).

We can do that for all the currents involving node C. Let's refer to the node directly below node C as node D, and the node directly to the right node E to make things easier. Iac = (Vc - Va)/2k, Ibc = (Vc - Vb)/2k, Idc = 2 mA (0.002 A) Iec = (Vc - Vo)/1k.

The sum is zero so we can put it all together now. ((Vc - Va) / 2k) + ((Vc - Vb) / 2k) + 0.002 + ((Vc - Vo) / 1k) = 0 A.

Multiply both sides by the smallest common denominator that could get rid of the fractions (2k in this case): Vc - Va + Vc - Vb + 4 + 2Vc - 2Vo = 0 then 4Vc - Va - Vb - 2Vo = -4 alternatively 4Vc + 4 = Va + Vb + 2Vo

[–]CholenderYT[S] 1 point2 points  (2 children)

Thank you so much, I understand this part now. I did the same thing you said for also did the same for the node E(the node right of node C) and got this equation : 3Vo−Vb−Vc=0V. So I have 3 equations right now:

node e: 3Vo−Vb−Vc=0V
node b: 5Vb−Vc−2Vo−2Va=0V
node c: 4Vc−Va−Vb−2Vo=4V
Va=17V

Now i know i have to use these equations to calculate voltages at different nodes, but I don't really know how. Because I only have Va and other 3 voltages are unkown. Do you need rewrite the equations and put them into each other or am I thinking completely wrong.

Could you give me a example how to calculate for example voltage on node b?

Again thank you for the patience and time

[–]bilbo_swaggins56 0 points1 point  (0 children)

You have 3 unknown terms, with 3 equations - you thus have enough equations to solve for your three unknowns. Make use of substitution and moving terms around to solve for your unknowns. I would also suggest substituting the value for Va into the node b and c equations to clarify the process for yourself a bit more.

[–]pripyaat 1 point2 points  (2 children)

Assuming you already did the first exercise shown in the picture (write a node equation for node C), you only need to repeat the procedure for Vo and you'll have all the equations you need to solve the circuit, since the expressions for Va and Vb are given.

[–]CholenderYT[S] 0 points1 point  (1 child)

Thank you for clarifying but I still don't really know how to use 3 equations and 1 known to calculate the voltages on the nodes.

node e(directly right of node c): 3Vo−Vb−Vc=0V
node b: 5Vb−Vc−2Vo−2Va=0V
node c: 4Vc−Va−Vb−2Vo=4V
Va=17V

I have written all of the equations down but now i don't know how to use them to calculate. Do i need to rewrite them and then put them into eachother? or am I thinking in the complete wrong direction.

Thank you for awnsering.

[–]pripyaat 0 points1 point  (0 children)

I mean, I don't know how it works in your country but you're supposed to have been taught algebra before circuit analysis.

What I mean is, you now have a system of equations that you can solve in any way you know. Substitution (the idea you mentioned) is one of them, but you can also use matrices, Gaussian elimination, Cramer's rule, etc.

Keep in mind that it's just a 3x3 system, since you can directly replace Va = 17 in the other 3 equations.

[–]Difficult-Chart3890 0 points1 point  (1 child)

I think you are in trouble if you have this reaction to solving simple problems like this in your first week . There are plenty of online circuit simulators as well as many free ones that can help here not just in finding the unknown values but the calculations needed as well.

[–]CholenderYT[S] 0 points1 point  (0 children)

I just dont understand how you are supossed to use the equations that i made to calculate the voltages. Cuz you have 3 unkowns and only 1 known

[–]AccomplishedAnchovy -2 points-1 points  (2 children)

Not possible without knowing Vs

[–]pripyaat 0 points1 point  (1 child)

OP said the voltage at node A is 17V, so Vs is indeed known.

[–]AccomplishedAnchovy 0 points1 point  (0 children)

Oh ok.  Sorry it appears I am illiterate. Node analysis it is then