Analyzing a DC circuit

Question

How would I analyze this DC circuit? I need to find the nodes, the branches, and the loops. I need the equations describing the circuit based on Kirchoff's laws.

I determined that the circuit has 6 nodes, 8 branches, therefore 3 loops. R3, R5, R6 (I would choose this as the ground)

For each node (except the ground node), I apply KCL:

Node 1 (R1, V1, R2):
IR1=IR2
Node 2 (R2, R3, R4):
IR2 = IR3 + IR4 
Node 3 (R4, I1):
IR4 = I1
Node 4 (I1, R6):
I1 = IR6
Node 5 (R3, R5, R6) ground
Node 6 (R1, V1, R5):
IR5 = IR1

I'm lost with the loops though.

Answer 1

The loops are given (from left to right) by: L1: R1 and V1 L2: V1, R2, R3 and R5 L3: R3, R4, I1 and R6

You already applied KCL, now you can also apply KVL to those loops

Answer 2

First, kirchoff's law is not correct since in Node 1 you are missing V1 current. Node 1 could be:

IR1+IR2 = IV1.

Node 6 has the same problem.

Also, your statement:

has 6 nodes, 8 branches, therefore 3 loops.

Is not correct at all. You have more than three loops. I.e. R1, R2, R4, R5, R6 and I1 is also a closed loop you could use. Also, note that as it has already been said, if you properly define KCL equations you don't need KVL to solve your circuit, is a complete equation system you could solve.

Despite that, for the KVL (or the loops) what you know is that for every closed loop in your circuit, the sum of all the voltages drops is equal to 0. The easiest example in your circuit is the first loop, R1, V1. You know that because of KVL, R1 voltage drop is 1V. Now, this was straight forward because they are directly in parallel. With this Information, you can now calculate R1. You could apply KVL by knowing that the voltage drop in each resistor is Rx is Ix times Rx. By defining this you have a whole new. set of equations that you can use to calculate the current in each resistor.

Now, be aware that you need to respect current direction in each loop. You first have to define an arbitrary direction. You want to sum all voltages so you need to decide in which direction will do so. Once you have defined Ix in each resistor, you know "in what direction" the voltage drops. There for, while summing up all voltages, you have to take into account such direction to know if you have to add them up or subtract them.

Let me use loop one as an example. We have defined I1 from top to bottom. So if we read the loop in the direction we have defined by the inner arrows (note they are just drawn to represent how are we reading the loop) we can see we have:

1. +1V of V1
2. -I1*R1 of R1 (note that in the direction we are reading the loop, R1 voltage goes from higher to lower).

With this you have: 1-I1*R1=0 because of KVL (which we already knew because they are in parallel). you can now compute I1 since you know R1.

If we had defined the opposite direction, our equation would strictly be I1*R1 - 1 = 0, which is actually the same equation in the other side of the equality sign.

Answer 3

almost completed labeling

Just for clarity (your image is far too fuzzy) and to graphically label the nodes you identified in your writing:

^{simulate this circuit – Schematic created using CircuitLab}

ground reference

All of these nodes cannot be left as undefined values. Take the following circuit, for example:

^{simulate this circuit}

Start at the bottom and go clockwise, assuming the current is non-zero. If so, then it must be the case that \$V_b\gt V_a\$ due to Ohm's Law and \$R_1\$. But it must also be the case due to Ohm's Law and \$R_2\$ that \$V_a\gt V_b\$. These are contradictory statements.

So we find that the only allowable possibility is that the current must be zero. It cannot be otherwise.

But even so, what's the value of \$V_a=V_b\$? All answers are allowed. So there's an infinite number of choices. The only way to remove that ambiguity is to assign \$0\:\text{V}\$ to either \$V_a\$ or \$V_b\$. As soon as that's done then the other has a single defined value.

We call the node assigned \$0\:\text{V}\$ by many names. One of them is ground or ground reference. But it is important to have one. You can assign it anywhere you want. But you do need to assign \$0\:\text{V}\$ to one and only one node.

There's a larger reason that applies to all non-trivial circuits, making exactly this same requirement for all such circuits, and it's called called the Rank-Nullity Theorem. But that's for another time.

completed labeling

So one of the nodes must be selected as \$0\:\text{V}\$:

^{simulate this circuit}

But KCL is never performed for known-voltage (or ground) nodes. Only if the node voltage is unknown is it kept and not removed from analysis.

So the circuit becomes:

^{simulate this circuit}

But we aren't done, yet.

A current source has \$\infty\$ impedance. They also have two ends. But in this case, \$\infty\$ impedance means you can split it into two parts, with one part pointing into a node (and coming out of nowhere) and one pointing away from a node (and going nowhere.)

^{simulate this circuit}

The gap between the two currents is just another representation of \$\infty\$ impedance in the original current source.

From the above, you know that \$V_4=V_5+I_1\cdot R_6\$ and that \$V_3=V_2-I_1\cdot R_4\$.

At this point, you should be able to see that there are only two unknown node voltages. Everything else has been defined as a specific value or a specific value relative to an unknown.

So the KCL equations are:

$$\begin{align*} \frac{V_2}{R_2}+\frac{V_2}{R_3}+I_1&=\frac{+1\:\text{V}}{R_2}+\frac{V_5}{R_3} \\\\ \frac{V_5}{R_3}+\frac{V_5}{R_5}&=\frac{V_2}{R_3}+I_1 \end{align*}$$

Note that \$R_4\$ and \$R_6\$ aren't present. That's because they don't affect the current leaving \$N_2\$ or entering \$N_5\$.

Once you solve that pair of equations, you can then work out \$V_3\$ and \$V_4\$ from \$V_2\$ and \$V_5\$.

question

I need to find the nodes, the branches, and the loops. I need the equations describing the circuit based on Kirchoff's laws.

On first blush, this sounds like you want to analyze this using both mesh and nodal, separately. But directed graph theory combines both in a single descriptive structure. So that's would be the answer if you are looking for an approach that combines both. (It can also readily show you the meshes.) You can look at my answer here to see if that's where you wanted to go. But I don't want to compose this problem in that fashion unless you say that's the goal. So I'll leave it here, for now.

Answer 4

First, label everything:

^{simulate this circuit – Schematic created using CircuitLab}

Ensure that all resistor voltages are labelled in accordance with passive sign convention. That is, the positive sign goes at the end where current enters the resistor. This isn't strictly necessary, but doing otherwise will likely lead to confusion with signs later on.

There are only 4 nodes where current branches, so called "branch nodes", labelled A, B, C and D. These are where you must apply KCL:

$$ \begin{aligned} I_4 - I_8 - I_2 &= 0 \\ \\ I_8 + I_5 - I_4 &= 0 \\ \\ I_2 - I_1 - I_3 &= 0 \\ \\ I_1 + I_3 - I_5 &= 0 \\ \\ \end{aligned} $$

There are three loops, and these are where we apply KVL. I will go clockwise around all three. As I go around a loop, for each element that I traverse I will add voltage when that traversal incurs a rise in potential, and I will subtract whenever I encounter a fall in potential, in keeping with my annotated voltage polarities:

$$ \begin{aligned} +V_8 - V_1 &= 0 \\ \\ +V_1 - V_2 - V_3 - V_5 &= 0 \\ \\ +V_3 -V_4 - V_7 - V_6 &= 0 \\ \\ \end{aligned} $$

Usually we can combine Ohm's law and KVL into a single step, but I will separate them here, and combine them in a moment. This is Ohm's law applied to all the resistors:

$$ \begin{aligned} V_8 &= I_8R_1 \\ \\ V_2 &= I_2R_2 \\ \\ V_3 &= I_3R_3 \\ \\ V_5 &= I_5R_5 \\ \\ V_4 &= I_1R_4 \\ \\ V_6 &= I_1R_6 \\ \\ \end{aligned} $$

Plugging these voltages into the KVL equations will reduce them all (with the exception of \$V_7\$) to be in terms of currents \$I_X\$, something that we would usually already have done while writing those KVL equations:

$$ \begin{aligned} I_8R_1 - V_1 &= 0 \\ \\ 1 - I_2R_2 - I_3R_3 - I_5R_5 &= 0 \\ \\ I_3R_3 - I_1R_4 - V_7 - I_1R_6 &= 0 \\ \\ \end{aligned} $$

These are all the simultaneous equations collated, that we must solve:

$$ \begin{aligned} I_4 - I_8 - I_2 &= 0 \\ \\ I_8 + I_5 - I_4 &= 0 \\ \\ I_2 - I_1 - I_3 &= 0 \\ \\ I_1 + I_3 - I_5 &= 0 \\ \\ I_8R_1 - V_1 &= 0 \\ \\ 1 - I_2R_2 - I_3R_3 - I_5R_5 &= 0 \\ \\ I_3R_3 - I_1R_4 - V_7 - I_1R_6 &= 0 \\ \\ \end{aligned} $$

While these equations form a complete description of the state of the circuit, and you can solve them to find all unknowns, there are a couple of interesting points to note.

A quick check will show that there are six unknown quantities, \$I_4\$, \$I_8\$, \$I_2\$, \$I_5\$, \$I_3\$ and \$V_7\$, but seven equations. The extra one is due to \$I_2=I_5\$, which can be deduced intuitively, if we encase the entire right side in a "black box":

^{simulate this circuit}

With nowhere else for current to enter or leave that dotted box, by KCL it's already clear that current entering must equal current leaving via the only two paths available, thus \$I_2=I_5\$.

Another observation is that current \$I_8\$ can be evaluated immediately:

$$ \begin{aligned} I_8R_1 - V_1 &= 0 \\ \\ I_8 &= \frac{V_1}{R_1} \\ \\ &= \frac{1V}{10\Omega} \\ \\ &= 100mA \\ \\ \end{aligned} $$

A similar situation will arise every time you have an independent voltage source directly across a resistor. In fact, \$I_8\$ could be disregarded completely, as you will find that it will disappear from the equations as you try to solve them. Even with R1 removed, the following will yield exactly the same results for the remaining currents and potentials:

^{simulate this circuit}