# kirchhoff laws with current source connected in series to resistor

I'm struggling with the following circuit trying, using kirchhoff laws, to write equations system that solve all variables. simulate this circuit – Schematic created using CircuitLab

(Ground symbols are used as arrows to show current direction)

I'm using source convention for sources and the other one for passive.

What I did so far is to define kirchhoff laws for each net clockwise

Voltages $$Net1(ABD) \Rightarrow V_{ab}-V_4=5$$ $$Net2(ACD) \Rightarrow -V_3+V_5=-5$$ $$Net3(BC) \Rightarrow V_4-V_5-V_6=0$$

Currents $$Net1(ABD) \Rightarrow I_2-I_3=-10$$ $$Net2(ACD) \Rightarrow I_4-I_6=10$$ $$Net3(BC) \Rightarrow I_3+I_5+I_6=0$$

So now problems come, writing single characteristics: $$V_{CA}=R_3\cdot I_3$$ $$V_{BD}=R_4\cdot I_4$$ $$V_{CB}=E6-(R_6\cdot I_6) \Rightarrow V_{CB}+(R_6\cdot I_6)=10$$

My problem is: how can I write the characteristic for $V_{AB}$??

• There's no real system to your approach and some of what I can fathom in your equations aren't even right (though, of course, you can define polarities and directions arbitrarily and perhaps escape the accusation.) Looks "hunt and peck" to me, like a chicken picking up bits at random. Bottom line is that I don't even know how to start to help you. Have you been taught mesh (KVL-like)? Or nodal (KCL-like)? Or? – jonk Oct 21 '17 at 19:23
• @jonk For what I have seen on the uni materials the first three equations I defined are KVL, the second three are KCL and at the end I add the equations of the bipoles. Then I put these equations inside a matrix and I get the answers. Am I completely wrong? If so, It would be great and I would appreciate if you could show me how to solve this problem. – LPs Oct 21 '17 at 19:32
• It's best to stick to a system: KCL or KVL. I think KCL is better/easier to use as a rigorous system approach as it doesn't require the recognition of, and sorting out of, appropriate "loop currents." It just focuses on nodes, which are easier to parse in a consistent fashion without making mistakes. It's also the approach used in Spice. But if you aren't familiar with nodal, applying it here might not really help but instead confuse. – jonk Oct 21 '17 at 19:44
• I've added an approach. Hopefully, you find it useful. – jonk Oct 21 '17 at 21:53

So here is your schematic without all the weird arrows that you added: simulate this circuit – Schematic created using CircuitLab

First step is to pick a node and call it zero (ground.) You get to do this with exactly one node of any schematic. If I see a node with LOTS of branches ending in it, I often will pick that one as ground as it reduces the number of terms in some of the expressions. But it really doesn't matter in the end.

Given how you assigned node names, I've selected this ground: simulate this circuit

Second step is to dump $R_1$. It serves no purpose. A current source has infinite impedance and any finite resistance in series with it is entirely irrelevant. The only purpose it serves is to change the current source's compliance voltage (the change of which of course is immediately dropped by the change in the resistance.)

So just dump it. simulate this circuit

Third step is to label the remaining nodes. simulate this circuit

Given that we've both arrived at the same number of nodes, I suspect you already have the skills to apply what I did above. So that's good.

Now, as the fourth step, just write out equations for each of the labeled nodes:

\begin{align*} \frac{V_A}{R_3} &= \frac{V_C}{R_3} + I_{E_2} + I_{A_1}\tag{V_A}\\\\ \frac{V_B}{R_4} + \frac{V_B}{R_6} + I_{A_1} &= \frac{V_D}{R_4} + \frac{0\:\textrm{V}}{R_6}\tag{V_B}\\\\ \frac{V_C}{R_3} + \frac{V_C}{R_5} &= \frac{V_A}{R_3} + \frac{V_D}{R_5} + I_{E_6}\tag{V_C}\\\\ \frac{V_D}{R_4} + \frac{V_D}{R_5} + I_{E_2} &= \frac{V_B}{R_4} + \frac{V_C}{R_5}\tag{V_D} \end{align*}

This is done by keeping all the outgoing currents on the left side and all of the incoming currents on the right side. Let's discuss this jonk's modified nodal method that you will not often see, but is really a whole lot easier to keep track of.

For each node, "imagine" that you are sitting in the middle of the node (which is like a flat, small, square floor sitting at some elevation you don't yet know on a long pole from "down there somewhere.") Water is spilling from above onto your floor. You don't know from how high, but you can tell it's spilling down onto your floor. Also, the water that hits your small floor now flows off the edges and spills down onto other floors below yours. You don't know where those are at, either. But you know they must be there. (The "water" is current, the heights of these small floors/nodes is the voltage of the floor/node.)

So the left side represents the currents spilling away and off the edges of your floor and onto other nearby floors. And the right side represents the currents spilling inward and onto your floor from other nearby floors.

(In keeping with this mental model, you can imagine that current and voltage sources may be like conveyors that may move water back upward to higher floors, so to speak. Also, no water is ever lost and all of it remains in play and the water cannot accumulate on the floors -- the net flow onto a floor must equal the net flow off of it -- KCL! -- and hence the justification for setting the left side equal to the right side.)

I've chosen a "direction" for $I_{E_2}$ and $I_{E_6}$, made evident by whether I placed that current on the left or right side of an equation. The only rule is that I am consistent about it. Here, I've decided that the current flows out of the (+) terminal of a voltage source.

Now, it's at this point that it would appear that there are six unknowns, $V_A$, $V_B$, $V_C$, $V_D$, $I_{E_2}$, and $I_{E_6}$, and only four equations. One way to resolve this is to have used so-called "supernodes" in the above equation development. But that just introduces another bit of terminology I'd like to avoid here. So let's stick with the above equations.

What else do we know? How about the fact that $V_A=V_D+V_{E_2}$ and that $V_C=V_{E_6}$? Nice. Now we have six unknowns and six equations. So this works out. We can either substitute in these two equations into the other four, reducing the number of variables to four, or we can keep six unknowns and pile on these two added equations. How you approach this is up to you. But let's perform the substitutions since it yields fewer equations and it's easy to do without much chance for misunderstanding:

\begin{align*} \frac{V_D+V_{E_2}}{R_3} &= \frac{V_{E_6}}{R_3} + I_{E_2} + I_{A_1}\tag{V_A}\\\\ \frac{V_B}{R_4} + \frac{V_B}{R_6} + I_{A_1} &= \frac{V_D}{R_4}\tag{V_B}\\\\ \frac{V_{E_6}}{R_3} + \frac{V_{E_6}}{R_5} &= \frac{V_D+V_{E_2}}{R_3} + \frac{V_D}{R_5} + I_{E_6}\tag{V_C}\\\\ \frac{V_D}{R_4} + \frac{V_D}{R_5} + I_{E_2} &= \frac{V_B}{R_4} + \frac{V_{E_6}}{R_5}\tag{V_D} \end{align*}

Now the unknowns are just $V_B$, $V_D$, $I_{E_2}$, and $I_{E_6}$.

The matrix solution is then:

\begin{align*} \begin{bmatrix}V_B \\ V_D \\ I_{E_2}\\I_{E_6}\end{bmatrix} &=\begin{bmatrix}0 & \frac{1}{R_3} & -1 & 0 \\ \frac{1}{R_4}+\frac{1}{R_6} & \frac{-1}{R_4} & 0 & 0 \\ 0 & \frac{-1}{R_3}+\frac{-1}{R_5} & 0 & -1 \\ \frac{-1}{R_4} & \frac{1}{R_4}+\frac{1}{R_5} & 1 & 0\end{bmatrix}^{-1} \begin{bmatrix}\frac{V_{E_6}-V_{E_2}}{R_3} + I_{A_1} \\ - I_{A_1} \\ \frac{V_{E_2}-V_{E_6}}{R_3} - \frac{V_{E_6}}{R_5}\\\frac{V_{E_6}}{R_5}\end{bmatrix} \end{align*}

If you stick those equations into some solver, or do it manually, you should get:

\begin{align*} V_B &=-\frac{5}{8}\:\textrm{V}\\ V_D &=8\frac{3}{4}\:\textrm{V}\\ I_{E_2} &=-8\frac{1}{8}\:\textrm{A}\\ I_{E_6} &=-\frac{5}{8}\:\textrm{A} \end{align*}

And from there you should be able to answer any quantitative questions about the rest of the circuit, such as the value for $V_A-V_B$.

• I really appreciate you efforts. Tomorrow morning I'll try to understand all you wrote. Thank you very much. – LPs Oct 21 '17 at 21:55
• @LPs Best wishes. Ask questions where you have them. – jonk Oct 21 '17 at 22:05
• Thank you. I understand you solution, but I cannot use it in university because of I need a matrix based solution. At the end I got it using the (as we call it in Italy) Net currents method and I posted it below. Thank you very much for all your efforts. – LPs Oct 23 '17 at 11:48

Thanks to @jonk to its answer that was very helpful, but for study purpose I need a matrix based solution and at the end I got it using the (as we call it in Italy) Net currents method

Net equations

\begin{equation} \begin{cases} \text{N1: } V_1-E_2-V_4=0 \newline \\ \text{N2: } E_2-V_3+V_5=0 \newline \\ \text{N3: } V_4-V_5-E_6+V_6=0 \end{cases} \end{equation}

That is

\begin{equation} \begin{cases} \text{N1: } V_1-V_4=E_2 \newline \\ \text{N2: } -V_3+V_5=-E_2 \newline \\ \text{N3: } V_4-V_5+V_6=E_6\\ \end{cases} \end{equation}

Bipoles characteristics are

• $V_3=R_3\cdot I_3$
• $V_4=R_4\cdot I_4$
• $V_5=R_5\cdot I_5$
• $V_6=R_6\cdot I_6$

Nets currents are

• $I_1=J_1$
• $I_2=J_2-J_1$
• $I_3=J_2$
• $I_4=J_1-J_3$
• $I_5=J_3-J_2$
• $I_6=-J_3$

So substituting characteristics into equations

\begin{equation} \begin{cases} V_1-(R_4\cdot I_4)=E_2 \newline \\ -(R_3\cdot I_3)+(R_5\cdot I_5)=-E_2 \newline \\ (R_4\cdot I_4)-(R_5\cdot I_5)+(R_6\cdot I_6)=E_6 \end{cases} \end{equation}

Substituting net currents

\begin{equation} \begin{cases} V_1-(R_4\cdot (J_1-J_3))=E_2 \newline \\ -(R_3\cdot J_2)+(R_5\cdot (J_3-J_2)=-E_2 \newline \\ (R_4\cdot (J_1-J_3))-(R_5\cdot (J_3-J_2))+(R_6\cdot -J_3)=E_6 \end{cases} \end{equation}

That is

\begin{equation} \begin{cases} -R_4J_1 +R_4J_3 +V_1=E_2 \newline \\ J_2(-R_3-R_5)+R_5J_3=-E_2 \newline \\ R_4J_1+R_5J_2+J_3(-R_4--R_5-R_6)=E_6 \end{cases} \end{equation}

Adding a 4th equation $J_1=A_1$ the system is

\begin{equation} \begin{cases} -R_4J_1 +R_4J_3 +V_1=E_2 \newline \\ J_2(-R_3-R_5)+R_5J_3=-E_2 \newline \\ R_4J_1+R_5J_2+J_3(-R_4--R_5-R_6)=E_6 \newline \\ J_1=A_1 \end{cases} \end{equation}

So solution will be retrieved using the following matrices

$$\begin{bmatrix} J_1\\ J_2 \\ J_3\\ V_1 \end{bmatrix}= \begin{bmatrix} -R_4 & 0 & R_4 & 1 \\ 0 & (-R_3-R_5) & R_5 & 0 \\ R_4 & R_5 & (-R_4-R_5-R_6) & 0\\ 1 & 0 & 0 & 0 \end{bmatrix}^{-1}\cdot \begin{bmatrix} E_2\\ -E_2 \\ E_6\\ A_1 \end{bmatrix}$$

• Well, I used a matrix to solve mine, as well. That's how I did it. But you need to approach your homework the way you are asked. So glad to hear you got it worked out. I've included my matrix in my solution so that you can see it. – jonk Oct 23 '17 at 16:39