How do we solve the circuits shown in Fig 1 using either node equation technique or loop equation technique? I am asking this question because source transformation cannot be applied to these circuits. Is there any other formal method to get the solution?
Dear Anup, I think we can't.
My conjecture is: Circuits which can be solved neither by node equation technique nor by loop equation technique have a "trivial" solution in the sense that the solution results from repeated application of Ohm's and Kirchhoff's laws.
This is certainly the case with your examples. Do you know of counter-examples to my conjecture?
Dear Joerg,
Could you please explain what do you mean by "trivial" solution. Is it all the variables must have zero values? It is not so for these examples. Moreover, What is meant by " solution results from repeated application of Ohm's and Kirchhoff's laws"? There must be a formal algorithm for that. I am looking for a formal method.
The solving is not "either node or loop" - it's "node and loop"!
While the left image is not complete (necessary for the loop equation), ground connections in both images are superfluous.
Solving the right circuit might be somewhat 'tricky' as the constant current (?) I' would imply a variable resistance. Anyway, Kirchhoff's laws still apply.
The left circuit also contains a current source that can be viewed when you click on the image. The word "tricky" implies lack of formalism. My question is can we get a formal method for this class of circuits. The class is defined as a circuit containing both ideal current and voltage sources and linear resistances.
OK - got that.
Assuming ideal V and I sources ...
The solution for the loop with the voltage source is quite trivial:
V = R * I3 (in both images).
In the left image I == I1, in the right image I == I2. ('==': identity)
This all with loop and line rules.
Add the rule for the nodes and the remaining variables are easily calculated.
How do you get V = R* I3 ?. What would be the contribution of the current source on I3 ?
How do you get I = I2 for the right image? We should apply Kirchoff's law at a particular node. What would happen if we have a parallel resistance or an ideal current source across the rightmost voltage source?
The problem becomes more complicated when we extend our definition of the assumed class by including controlled sources.
I'm somewhat late due to limitations during the day ...
Basic reference: https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws
KVL: "The (vector) sum of the voltages in a loop is 0."
==> (Both images) V - I3*R = 0 ==> V = I3*R
(Slightly modified) KCL: "The current in the connection between 2 nodes is constant."
(You are free to define a node anywhere in that connection. KCL: "The (vector) sum of currents into a node is 0.")
Thus (left image, wire forming most of the left loop): I = I2
(Left image) The effective contribution of the constant current on I3 is 0 (zero): this current will have flow through the voltage source (in opposite direction of what you'd expect). Not a big issue as the (ideal) voltage source's source resistance is 0.
The right image poses a bigger challenge as the current source has to be somewhat counter-intuitive: in the lower loop both voltages are in series. Thus the current source has to act like some controlled resistance. Not impossible, just unusual.
All the above analysis is based on ideal voltage resp. current sources.
"What will happen if ..." ?
If you add some parallel resistance, current source etc., you introduce additional loops and nodes. New problem :)
(But if it can be solved at all, it is solvable: the *Spice simulation programs rely massively on KCL and KVL - building and solving matrices.)
I admit that I made a mistake in commenting on the application of KVL, that is ,
V=R*I3 .
I could not understand the modified KCL part that is " The current in the connection between 2 nodes is constant." Are the nodes are distinct electrical nodes as in the case of Fig 1(b)? Have you applied this rule to write I == I2. ? Is it a general rule? Where can I get it?
I think for the left image the complete solution is
I=I1
I1=I2 + I3
and I3 = V/R
Similarly for the right image
I3 = V/R
and I = I1 + I3
I1 + I3 = I2
Am I correct?
Oops - the last post should have been I = I1 for the left image (just as in the first post with equations, where I also stated I = I2 for the right image).
The constant current on a wire between 2 nodes is a "shortcut" (you may call it "rule") that can be derived from the KCL.
Your equations seem to be correct.
Dear Anup,
Dear Colleagues,
Welcome,
It is nice to meet,
I would tend to adopt the opinion of the colleague Dreher that one can use a mixed loop and node equations to reach to the solution in the shortest way.
The key start is to get I3 as Dreher said. V= I3 R in both circuits as the voltage across R is clamped by the voltage source. This can be considered as a loop equation.Then we can get I1 in circuit b by mode equation I1= I- I3, then one gets I2 by node equation I2=I3+ I1.
As for the circuit a: I2= I1-I3 as node.
It is clear that the circuit contains a mix of current and voltage source then one can use node and loop and node equations.
One can use also the electric circuit theories to solve these circuits. Here one can use the superposition theory.
Best wishes
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