Application of the fundamental laws
In circuit design and analysis the “fundamental laws” are,
- Ohm’s Law for resistors
- The corresponding $i$-$v$ equations for capacitors and inductors
- Kirchhoff’s Laws
We used Ohm’s Law a few times already to figure out series and parallel resistors, the voltage divider, and simplifying a resistor network. Now we add Kirchhoff’s Current Law and Kirchhoff’s Voltage Law to our bag of tricks for solving circuits.
Contents
- Apply the fundamental laws
- Assign names and polarity
- Select the independent variable
- Write independent equations
- Solve the system of equations
- Solve the other variables
Where we’re headed
Ohm’s Law in combination with Kirchhoff’s Laws (KCL and KVL) are powerful tools for solving simple circuits, using this strategy,
- Give names to voltages and currents.
- Select the independent variable, either $i$ or $v$.
- Write a set of independent KCL and KVL equations.
- Solve the system of equations.
- Solve for the other unknown currents or voltages.
Our task is to find the unknown currents and voltages in this circuit,
Apply the fundamental laws
We will take these steps to work our way to a solution,
- Assign names and polarity to voltages and currents using the sign convention for passive components.
- Select the independent variable to produce the simplest equations. We have a choice between $i$’s or $v$’s as the independent variable.
- Write equations using KCL, KVL, or both. Make sure every circuit element participates in at least one equation.
- Solve the system of equations for the independent variable.
- Solve the remaining unknowns.
Assign names and polarity
The first step in a circuit analysis is to give symbolic names to voltages, currents, and nodes. This creates our vocabulary for writing equations.
Let’s make a list of what we know about this circuit, the features of the circuit,
- $5$ circuit elements
- $3$ nodes, $\green a$, $\green b$, and $\green c$
- $3$ meshes (inner loops)
- $1$ source voltage, $v_{\text S}$
- $1$ source current, $i_{\text S}$
and another list of what we don’t know, yet, the voltages and currents we will discover during the analysis,
- $2$ element voltages, $v_1$ and $v_2$
- $3$ element currents, $i_1$, $i_2$ and $i_3$
review node, loop, mesh
A node is a junction where $2$ or more elements connect.
A loop is any closed path through circuit elements.
A mesh is a loop that has no other loops inside it.
When we assign polarity to voltage and current labels we use the sign convention for passive components. The current arrow points into the positive voltage of each resistor.
This circuit has three nodes, redrawn here to highlight the junctions at nodes $b$ and $c$.
(There is an obvious opportunity to simplify the two parallel resistors, $6\,\Omega$ with $5\,\Omega$. However, we will not do that, because we want to focus on the analysis procedure.)
Select the independent variable
At this point we have to (get to) choose the independent variable. Should it be $v$ or $i$?
A good way to make this choice is to compare the number of unknown voltages to unknown currents. There are $2$ unknown voltages, and $3$ unknown currents. If we select voltage as the independent variable, we will have equations with $2$ voltage terms as opposed to $3$ current terms. $2$ is less that $3$, so we choose voltage as our independent variable.
Write independent equations
Since we have two unknowns, $v_1$ and $v_2$, we need to come up with two independent equations.
I look at the circuit and pick two interesting features. Node $b$ has several connections, making it a focal point of the circuit. The left mesh nicely includes the all the remaining circuit elements that don’t already touch node $b$.
Our choice will be,
- a KVL equation around the left mesh
- a KCL equation at node $b$
Admittedly, I used my own design experience to make these choices. As you do more problems like this, you will build your intuition, too.
When you write equations, make sure every circuit element participates in at least one equation. Every element has to have a chance to influence the outcome.
KVL around the left mesh
The left mesh is the one with the orange circle.
We start at the lower left corner where you see the ground symbol, and travel clockwise around the mesh adding up voltages. Kirchhoff’s Voltage Law says the sum of element voltages around a loop must add up to zero.
$\displaystyle \sum_n v_n = 0$
$+v_{\text{S}} - v_1 - v_2 = 0$
We know $v_{\text S}$ is $140\,\text V$ by inspection—it appears directly across the voltage source,
$+140 - v_1 - v_2 = 0$
The $-$ signs for $v_1$ and $v_2$ are because we encounter their $+$ sign first during the clockwise tour around the loop. We see a voltage drop as we go through the component.
What is the big zig-zag symbol?
The $\sum$ symbol is the Greek letter capital Sigma. In mathematical notation it is the summation operation. It adds up a series of numbers. $v_1 + v_2 + v_3 + … + v_n$.
KCL at node $b$
We will get our second equation by writing Kirchhoff’s Current Law at node $b$.
One form of Kirchhoff’s Current Law says the currents flowing into a node must equal the currents flowing out of the node.
$\displaystyle \sum_n i_{in_n} = \sum_m i_{out_m}$
Add up the currents flowing into node $b$, set them equal to the sum of currents flowing out.
$i_1 + i_{\text S} = i_2 + i_3 $
Earlier in Step 2. we decided to use $v_1$ and $v_2$ as the independent variables, so we use Ohm’s Law to express the unknown currents in terms of voltage and resistance. We can fill in the value of $i_{\text S}$ immediately, since it comes directly from the current source.
$\dfrac{v_1}{20\,\Omega} + 18 = \dfrac{v_2}{6\,\Omega} + \dfrac{v_2}{5\,\Omega} $
After a little rearrangement, we have our system of two equations in two unknowns,
$v_1 + v_2 = 140$
$\dfrac{1}{20} v_1 - \left (\dfrac{1}{6} + \dfrac{1}{5}\right ) v_2 = -18$
These two equations capture everything going on in our circuit.
This is a good time to do a quick check. Did every circuit element get a chance to participate in at least one equation? Are any left out? Account for all $5$ elements.
Solve the system of equations
I encourage you to take the time to solve this problem yourself before looking at the answer. This system of equations can be solved as simultaneous equations—that’s how I’ll do it—or by linear algebra techniques like Cramer’s Rule.
Find unknown voltages $v_1$ and $v_2$.
$v_1 =$ _____ $\text V$
show $v_1$
$v_1 = 80\,\text V$
$v_1 =$ _____ $\text V$
show $v_2$
$v_2 = 60\,\text V$
Solve the other variables
Find unknown currents $i_1$, $i_2$, and $i_3$.
$i_1 =$ _____ $\text A$
show $i_1$
$i_1 = 4\,\text A$
$i_2 =$ _____ $\text A$
show $i_2$
$i_2 = 10\,\text A$
$i_3 =$ _____ $\text A$
show $i_3$
$i_3 = 12\,\text A$
show the full solution
Solve the system of equations
The system of equations is,
$v_1 + v_2 = 140$
$\dfrac{1}{20} v_1 - \left (\dfrac{1}{6} + \dfrac{1}{5}\right ) v_2 = -18$
Eliminate $v_2$ by solving the first equation for $v_2$,
$v_2 = 140 - v_1$
Then substitute for $v_2$ in the second equation,
$\dfrac{1}{20} v_1 - \left (\dfrac{1}{6} + \dfrac{1}{5}\right ) (140 - v_1) = -18$
Solve for $v_1$,
I carry the numbers along as fractions so there is no rounding error.
The least common multiple of $6$ and $5$ is $30$.
$\dfrac{1}{20} v_1 - \left (\dfrac{5}{30} + \dfrac{6}{30}\right ) (140 - v_1) = -18$
$\dfrac{1}{20} v_1 - \left (\dfrac{11}{30}\right ) 140 + \left (\dfrac{11}{30}\right ) v_1 = -18$
Put $v_1$’s on the left, and constants on the right,
$\dfrac{1}{20} v_1 + \dfrac{11}{30} v_1 = \left (\dfrac{11}{30}\right ) 140 - 18$
And crank the algebra,
$\left (\dfrac{3}{60} + \dfrac{22}{60} \right )v_1 = \dfrac{11 \cdot 140}{30} - \dfrac{540}{30}$
$\dfrac{25}{60} \,v_1 = \dfrac{1540}{30} - \dfrac{540}{30}$
$\dfrac{25}{60} \,v_1 = \dfrac{1000}{30} $
$v_1 = \dfrac{60}{25}\cdot\dfrac{1000}{30} $
$v_1 = 80\,\text V$
One voltage down, one to go,
$v_2 = 140 - v_1$
$v_2 = 140 - 80$
$v_2 = 60 \,\text V$
$v_1$ and $v_2$ are both solved.
Now we can go after the unknown currents, using Ohm’s Law.
Solve for the other unknowns
$i = \dfrac{v}{\text R}\qquad$ Ohm’s Law
$i_1 = \dfrac{v_1}{20} = \dfrac{80}{20} \qquad i_1 = 4\,\text A$
$i_2 = \dfrac{v_2}{6} = \dfrac{60}{6} \qquad i_2 = 10\,\text A$
$i_3 = \dfrac{v_2}{5} = \dfrac{60}{5} \qquad i_3 = 12\,\text A$
All done! The solved circuit looks like this,
Summary
We solved a circuit by direct application of the fundamental laws. Our tools were Ohm’s Law and Kirchhoff’s Laws.
The steps to a solution,
- Label voltages and currents using the sign convention for passive components.
- Select the independent variable, either $i$ or $v$ to produce the simplest equations.
- Write equations using KCL, KVL, or both. Make sure every element is represented in at least one equation.
- Solve the system of equations.
- Solve for the other unknowns.
You may be thinking this process is somewhat arbitrary and improvised. You are not mistaken. Coming next, we will talk about two efficient and well-organized methods for solving any circuit, the Node Voltage Method, and the Mesh Current Method.