As you study circuit analysis, it may seem like luck or coincidence that you get the right number of equations to solve. Luck is not involved. The analysis methods you are learning reliably capture the information needed to solve the circuit.

“Solving a circuit” means solving a system of simultaneous equations.

  • How many equations are required to solve a circuit?
  • Where do they come from?

Written by Willy McAllister.


Contents


Where we’re headed

Solving a circuit means finding the voltage and current for every element. Each element has an unknown $i$ and $v$. This means we need twice as many independent equations as there are elements in the circuit.

These equations come from,

  • Half come from the $i$-$v$ laws for each circuit element.
  • The other half come from,
    • $N - 1$ come from Kirchhoff’s Current Law, where $N$ is the number of nodes.
    • $E - (N - 1)$ come from Kirchhoff’s Voltage Law, where $E$ is the number of elements.

The results we talk about here are baked into the different circuit analysis methods,

How many independent equations are needed to solve a circuit?

This is the key question that determines the amount of effort required to perform a circuit analysis. As we learned in algebra class when solving simultaneous equations, the number of independent equations you need to solve a system is the same as the number of unknown variables. If you have a system with $10$ unknowns, you need $10$ equations to solve for the $10$ unknowns.

How many unknowns does a circuit have? Every two-terminal element contributes one unknown voltage and one unknown current. So $E$ elements contribute $2E$ unknowns.

A circuit with $E$ elements requires a system of $2E$ independent equations.

Concept check

As we discuss this idea in abstract terms, we will also use this real circuit example. If you want to review your understanding of circuit terminology, check here.

How many elements are there in the circuit?

$E = $ ______ elements.

show answer

This circuit has $E=5$ elements.

How many nodes does the circuit have?

$N = $ ______ nodes.

show answer

$N=3$ nodes.

How many loops does the circuit have?

______ loops.

show answer

$6$ loops. The $3$ loops $\goldD{\text{I}}$, $\goldD{\text{II}}$, and $\goldD{\text{III}}$ are called meshes. A mesh is a loop that contains no other loops. A mesh is also called an inner loop.

How many of those loops are meshes?

______ meshes.

show answer

Of the $6$ loops in the circuit, $3$ of them are meshes (also known as inner loops). The meshes are numbered $\goldD{\text{I}}$, $\goldD{\text{II}}$, and $\goldD{\text{III}}$.

How many equations do we need to solve this circuit?

_______ equations.

show answer

This circuit has $E = 5$ elements. We have to come up with $2E = 10$ independent equations to solve this circuit.

Where do the equations come from?

The equations come from two sources: the elements themselves, and how the elements are connected to each other.

These constraints determine the currents and voltages in a circuit,

  • $i$-$v$ element laws
  • Kirchhoff’s Current Law
  • Kirchhoff’s Voltage Law

The system of equations you write captures these constraints. Half of the equations come from the individual element laws. The other half come from either KCL and/or KVL. Kirchhoff’s Laws “know” how circuit elements interact with each other.

Half of the equations come from element laws

Imagine unconnected circuit components scattered about on the tabletop.

Each element has unknown current and voltage,

Each element brings along an $i$-$v$ equation. Think of each element as little chunk of math.

These $i$-$v$ relations represent $E$ independent equations. That’s half the required total.

What about capacitors and inductors?

This example circuit does not include capacitors or inductors. If it did, each capacitor or inductor would contribute one $i$-$v$ equation,

$i = \text C \,\dfrac{dv}{dt}\quad$ or $\quad v = \text L\,\dfrac{di}{dt}$

Half of the equations come from Kirchhoff’s laws

The remaining $E$ equations come from the constraints caused by the connections between elements. Circuit connections constrain the voltages and currents of individual elements to be certain values. An example of a constraint would be, “these two elements are in series, so their currents have to be the same.” We can develop $E$ connectivity equations using Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL).

Let’s say a circuit has $E$ elements and $N$ nodes.

Elements and branches are the same thing.

Element and branch mean just about the same thing in circuit analysis. Every element (resistor, capacitor, inductor, source, or other $2$-terminal component) contributes one branch to a circuit. Branch is the graph theory term for a path connecting two nodes. When talking about a circuit as a graph, it is common to use the terms node and branch to describe the junctions and paths in the graph. In this article, I will stick with $E$ as the name for the number of elements/branches.

Our example circuit has $E = 5$ elements (branches) and $N = 3$ nodes. We also know the circuit has $6$ loops, $3$ of which are meshes.

Show me the nodes

Show me the loops and meshes

This example circuit has $6$ loops. The $3$ loops $\goldD{\text{I}}$, $\goldD{\text{II}}$, and $\goldD{\text{III}}$ are called meshes. A mesh is a loop that contains no other loops.

Having $3$ nodes and $6$ loops is a lot of possibilities for coming up with $E=5$ more equations, but we have to be careful. The equations we generate have to be independent of each other.

What is an independent equation?

An equation is linearly independent if it can’t be derived by linear combinations other equations. Linear combinations are when you merge two equations by adding, subtracting, or multiplying by a constant. There is an example in the next paragraph showing how in a system of KCL equations, one of them is not independent because it can be derived from other equations.

How many independent equations come from KCL?

We can write KCL equations for every node in the circuit. This will give you $N$ equations. BUT, the $N$ KCL equations are not independent. One of them is redundant. There is always one dependent equation that does not contribute any new information, so it is not needed.

Please demonstrate linear dependence

Let's write all the KCL node equations for our example circuit and show the linear dependence.

KCL for node $\text a$: $\quad +i_1 -i_1 = 0$
KCL for node $\text b$: $\quad +i_1 - i_2 - i_3 +i_{\text S} = 0$
KCL for node $\text c$: $\quad -i_1 + i_2 + i_3 -i_{\text S} = 0$

Node $\text a$ is trivial because there is one current coming in, and one going out.

Node $\text b$'s KCL equation can be used to derive the equation for node $\text c$: multiply node $\text b$ equation by $-1$ and you get node $\text c$. That means either node $\text b$ or node $\text c$ equation is redundant, and can be left out of the system of equations.

With KCL we get credit for $N-1$ independent equations. In our example circuit we have $3$ nodes, so KCL gives us $3-1=2$ KCL equations.

The node we leave out is a choice. Usually, we leave out the ground node because it is the most complicated (has the most connections). Here’s our equation-finding status up to now,

$\begin{aligned} &\text{Needed}: &2E \\ &\text{From }i\text - v: &E \\ &\text{From KCL}: &N-1 \end{aligned}$

This leaves $E - (N-1)$ equations we have to find using KVL.

How many independent equations come from KVL?

After writing $N-1$ equations using KCL, we come up short of $2E$ equations by $E - (N-1)$. For our example circuit, we need $5 - (3-1) = 3$ additional equations. Where will they come from? We use KVL around the loops of the circuit.

Graph theory tells us two wonderful things,

  • KVL produces exactly the right number of independent equations, $E - (N-1)$.
  • $E - (N-1)$ happens to be the same as the number of meshes.

That means we know the required number of KVL equations by counting the meshes.

Our example circuit has $3$ meshes, so we know immediately we need to write $3$ KVL equations; no more, no less.

Limitation: planar vs. non-planar circuits

KVL will produce $E - (N-1)$ equations for any kind of circuit, but we can only use the meshes if the circuit is planar.

A planar circuit is one that can be drawn flat, with no crossing wires. If a circuit cannot be drawn flat without crossing wires, it's non-planar. Our example circuit is planar, as are most of the circuits you will be asked to analyze by hand.

The Loop Current Method handles non-planar circuits.

Make sure the KVL equations are independent

We want the KVL equations to be independent. This requires a bit of care.

Simplest: Limit yourself to writing KVL equations for just the meshes. The meshes are guaranteed to produce the right number of equations, and they will be independent.

If for some reason we want to (or have to) include other loop equations, be sure the loop equations are independent. Here’s how: Make sure every loop includes one element not in any other loop. That will do the trick.

For our example circuit, we need to come up with $3$ independent KVL equations, selecting from the $6$ available loops.

The simplest way is to pick the three meshes, $\goldD{\text{I}}$, $\goldD{\text{II}}$, and $\goldD{\text{III}}$. We win! The meshes produce the right number of equations, and they are guaranteed to be independent. This is the basis of the Mesh Current Method.

We could choose another valid set of loops from our example circuit, $\greenD{\text{IV}}$, $\blueD{\text V}$, and $\maroonC{\text{VI}}$. Why might this be a good set?

  • There are $3$ equations, as required by $E-(N-1) = 3$.
  • Every element is included in a loop.
  • Every loop has an element not included in other loops.

Concept check

Some choices do not meet the guidelines. Can you tell why?

  • $\goldD{\text I}$, $\blueD{\text V}$, and $\maroonC{\text{VI}}\quad$
why?

This set misses the resistor next to the current source. Every element needs a chance to influence the result.

  • $\greenD{\text{IV}}$ and $\blueD{\text{V}}\quad$
why?

This set only produces $2$ equations, and $3$ are required. This is true, even though the loops together pass through every element. You still are required to have $3$ equations to fully describe/constrain the circuit. This is why the good set listed above works, $\greenD{\text{IV}}$, $\blueD{\text V}$, and $\maroonC{\text{VI}}$.

  • $\goldD{\text I}$, $\goldD{\text{II}}$, $\goldD{\text{III}}$, and $\maroonC{\text{VI}}\quad$
why?

You could get the answer with $4$ loops, but it's more than you need. $4$ exceeds the number of required equations, $E-(N-1) = 3$. That means one of the equations is linearly dependent on the others, and can be left out.

Don’t shy away from using loops; just be alert and thoughtful about it.

When might I want to select non-mesh loops? There are some circuits where we want to use loops, and others where we are forced to use loops in addition to meshes. These special cases are described in the Loop Current Method.

Summary

There are three constraints placed on currents and voltages in a circuit,

  • $i$-$v$ element laws
  • Kirchhoff’s Current Law
  • Kirchhoff’s Voltage Law

The system of equations you write captures these constraints.

For a circuit with $E$ elements and $N$ nodes,

  • You need
    • $2E$ independent equations to solve the circuit.
  • You get
    • $E$ equations from the $i$-$v$ law for each component (Ohm’s Law and the like).
    • $N-1$ independent node equations using KCL.
    • $E - (N-1)$ independent loop equations using KVL.

Writing KVL equations on meshes guarantees the right number of independent KVL equations.

If you write KVL equations for non-mesh loops, every loop has to have at least one element not in any other loop to guarantee independence.