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# Application of the fundamental laws

We solve a circuit by direct application of the fundamental laws: Apply element laws (Ohm's Law and the like) plus Kirchhoff's Laws to solve for the currents and voltages of a circuit. Written by Willy McAllister.

## Introduction

We've done a few examples of direct application of Ohm's Law when we derived equations for series and parallel resistors, a voltage divider, and simplifying a resistor network. Now we do an example that puts Kirchhoff's Laws to work for us, too. We call this the application of the fundamental laws.
Task: Find the unknown currents and voltages in this circuit.
The steps to a solution involve creating and solving a system of independent equations,
1. Label the voltages and currents using the sign convention for passive components.
2. Select the independent variable, either i or v to produce the simplest equations.
3. Write equations using KCL, KVL, or both. Make sure every element is represented in at least one equation.
4. Solve the system of equations.
5. Solve for any remaining unknown voltages and currents you want to know.

### Step 1: Label the schematic

To start, it helps to give names to voltages, currents, and nodes, and make a list of what we do and do not know.
Circuit features and unknowns:
• 5 elements
• 3 nodes, labeled start color #28ae7b, a, end color #28ae7b, start color #28ae7b, b, end color #28ae7b, and start color #28ae7b, c, end color #28ae7b.
• 3 meshes (inner loops).
• 1 source voltage, v, start subscript, start text, S, end text, end subscript, and 2 element voltages, v, start subscript, 1, end subscript and v, start subscript, 2, end subscript.
• 1 source current, i, start subscript, start text, S, end text, end subscript, and 3 element currents, i, start subscript, 1, end subscript, i, start subscript, 2, end subscript and i, start subscript, 3, end subscript.
When assigning polarity to the voltage and current of each element, we use the sign convention for passive components : The current arrow points into the positive voltage end of each resistor.
To emphasize there are only three nodes in this circuit, it is redrawn here to highlight the junctions at nodes b and c.
(There is an obvious opportunity here to simplify the two parallel resistors, 6, \Omega with 5, \Omega. We will not do that, because we want to study the general analysis procedure.)

### Step 2. Select the independent variable

At this point we have a choice to make. Should the independent variable be voltage v or current i? One 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 simpler, so voltage will be the independent variable for this problem.

### Step 3. Write independent equations

Since we have two unknown voltages, we need two independent equations to solve for them. Our choice will be a KVL equation around the left-most mesh and a KCL equation at node start color #28ae7b, b, end color #28ae7b. Why did I make these particular choices? I picked the two most interesting features of the circuit. Node start color #28ae7b, b, end color #28ae7b has several connections, making it an interesting focal point for the circuit, and the left-most mesh nicely includes the all the remaining circuit elements not fully controlled by node start color #28ae7b, b, end color #28ae7b. Admittedly, I used some of my own experience in electronics to anticipate the direction the analysis will take. As you do more problems of this sort, you will build your intuition, too.

#### KVL around the left-most mesh

The left-most mesh is the one with the orange circle.
We start at the lower left corner of the circuit, where you see the ground symbol, and go 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.
plus, v, start subscript, start text, S, end text, end subscript, minus, v, start subscript, 1, end subscript, minus, v, start subscript, 2, end subscript, equals, 0
plus, 140, minus, v, start subscript, 1, end subscript, minus, v, start subscript, 2, end subscript, equals, 0
The minus signs for v, start subscript, 1, end subscript and v, start subscript, 2, end subscript are because we encounter their plus sign first during the clockwise tour around the loop, indicating we will see a voltage drop as we go through the component.

#### KCL at node $\green b$start color #28ae7b, b, end color #28ae7b

We will get our second equation by writing Kirchhoff's Current Law at node start color #28ae7b, b, end color #28ae7b. One form of Kirchhoff's Current Law says the currents flowing into a node must equal the currents flowing out of the node.
Add up the currents flowing into node start color #28ae7b, b, end color #28ae7b, set them equal to the sum of currents flowing out.
i, start subscript, 1, end subscript, plus, i, start subscript, start text, S, end text, end subscript, equals, i, start subscript, 2, end subscript, plus, i, start subscript, 3, end subscript
Earlier in Step 2. we decided to use v, start subscript, 1, end subscript and v, start subscript, 2, end subscript as the independent variables, so we use Ohm's Law to express the unknown currents in terms of voltage and resistance.
start fraction, v, start subscript, 1, end subscript, divided by, 20, \Omega, end fraction, plus, 18, equals, start fraction, v, start subscript, 2, end subscript, divided by, 6, \Omega, end fraction, plus, start fraction, v, start subscript, 2, end subscript, divided by, 5, \Omega, end fraction
After a little rearrangement, we have our system of two equations in two unknowns,
v, start subscript, 1, end subscript, plus, v, start subscript, 2, end subscript, equals, 140
start fraction, 1, divided by, 20, end fraction, v, start subscript, 1, end subscript, minus, left parenthesis, start fraction, 1, divided by, 6, end fraction, plus, start fraction, 1, divided by, 5, end fraction, right parenthesis, v, start subscript, 2, end subscript, equals, minus, 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.

### Steps 4 and 5 - Solve

Have a go at solving this system of equations yourself before looking at the full answer below.
Find unknown voltages v, start subscript, 1, end subscript and v, start subscript, 2, end subscript, and unknown currents i, start subscript, 1, end subscript, i, start subscript, 2, end subscript, and i, start subscript, 3, end subscript.
v, start subscript, 1, end subscript, equals
start text, V, end text

v, start subscript, 2, end subscript, equals
start text, V, end text

i, start subscript, 1, end subscript, equals
start text, A, end text

i, start subscript, 2, end subscript, equals
start text, A, end text

i, start subscript, 3, end subscript, equals
start text, A, end text

## 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:
1. Label voltages and currents using the sign convention for passive components.
2. Select the independent variable, either i or v to produce the simplest equations. Choose the variable with the fewest unknowns.
3. Write equations using KCL, KVL, or both. Make sure every element is represented in at least one equation.
4. Solve the system of equations for the independent variables (in this case, v, start subscript, 1, end subscript and v, start subscript, 2, end subscript).
5. Solve for the other unknowns.

Our approach to solving this circuit was solidly based on the fundamental laws, and we got the right answer. But our choice of equations felt somewhat arbitrary. Coming up next, we will talk about two efficient and well-organized methods for solving any circuit, the Node Voltage Method, and the Mesh Current Method.

## Want to join the conversation?

• At the point where KCL is calculated at node b , why is the voltage same for currents i2 and i3 ? • i2 flows through the 6-ohm resistor. i3 flows through the 5-ohm resistor. The terminals of both of these resistors are connected together (both top terminals connect to node b, and both bottom terminals connect to node c). That means they have the same voltage. The voltage has been named v2. (The "2" in the voltage name is not related to the "2" in the i2 name.)
• How is it that current flows "up" through the current source in this example, from lower voltage to higher voltage? What am I missing here? Is it magic? • The current source in this circuit creates a current flowing up from node "c" to node "b". Like a battery, a current source is a power generator. Something is going on inside the current source that makes it able to force current to flow "up" against an increasing voltage. (A battery can do the same thing, it forces current to flow out its positive terminal because of a chemical reaction going on inside.) Voltage sources (batteries) are familiar to us, since we can go to the store and buy them. Current sources are not made from a chemical reaction. They are some sort of complicated electronic circuit that you don't run across at the grocery store. They are unfamiliar, but not magic. Pretty much every analog integrated circuit has circuits that act like a current source over a limited range of voltage.
• Why isn't there a voltage over the 5 ohm resistor? • excuse-me, at the lesson " step3- KCL at node B" <<-- how and where to get that i1 + is = i2 + i3;
(1 vote) • The KCL equation for node b is: ( i1 + is = i2 + i3 ). This is the form of Kirchhoff's Current Law where all the currents flowing into the node ( i1 + is ) are set equal to all the currents flowing out of the node ( i2 + i3 ). The blue current arrow for i1 is drawn on the left side of the 20-ohm resistor, but that same current flows out the right end of the resistor and goes into node b. So node b has two currents flowing in, and two flowing out.
• Please, could any of you recommend me web site for practising such exercises
(1 vote) • Here's a suggestion: This article and the following videos/articles on Node Voltage and Mesh Current methods all have example circuits. For each different circuit, see if you can solve them with more than one method.

Another very realistic way to practice is to make up simple circuits on your own and analyze them by hand. To check your work, draw your circuit in this simulator and run a DC analysis. http://spinningnumbers.org/a/circuit-sandbox.html

This simulator is part of a web site I've been working on since my EE Fellowship at KA completed. The articles are all updated and improved, and circuit simulation is incorporated as part of the learning process.
• Can you explain more about the current divider rule • please can i get help with how the line that follows "and crank the algebra" in the solution came about.
(1 vote) • How would one label the current flow in node c? I see that it is unimportant for solving the circuit, but I can't yet wrap my head around why. Particularly, I don't know how I would label the current on the node between the voltage source and the 6 ohm resistor, and then the branch between the two resistors (on node c).
(1 vote) • by my calculations the Vdrop over R1 is equal to:

( ( Vss / (R1 + ( ( R2 * R3 ) / ( R2 + R3) ) ) * R1

( ( 140v / ( 20 + ( ( 6*5 ) / ( 6 / 5) ) ) ) * 20

Rtot = R1 + Rp

Itot (voltage side) = Vss / Rtot

Vdrop over R1 = Itot (voltage side) * R1

And this results in a Vdrop over R1 equal to 123,18 V
This is ofcourse unless you count the "Current Source" as a Voltage source also which would interfere with the "voltage source given".
Am i totaly wrong here?
(1 vote) • Here's a good way to check your work. Simulate the circuit by copying this entire URL into a web browser. Then tap on *DC* to run a simulation. Compare your calculations to what the simulator thinks is happening.

http://spinningnumbers.org/circuit-sandbox/index.html?value=[["v",[104,96,0],{"name":"VS","value":"dc(140)","json":0},["2","0"]],["i",[320,144,6],{"name":"IS","value":"dc(18)","json":1},["0","1"]],["r",[120,72,3],{"name":"R1","r":"20","json":2},["2","1"]],["r",[200,96,0],{"name":"R2","r":"6","json":3},["1","0"]],["r",[256,96,0],{"name":"R3","r":"5","json":4},["1","0"]],["w",[104,96,104,72]],["w",[104,72,120,72]],["w",[200,72,200,96]],["w",[200,72,256,72]],["w",[256,72,256,96]],["w",[256,72,320,72]],["w",[320,72,320,96]],["w",[320,144,256,144]],["w",[200,144,256,144]],["w",[200,144,104,144]],["g",[200,144,0],{"json":15},["0"]],["w",[168,72,200,72]],["view",7.200000000000003,-5.259999999999998,1.953125,"50","10","1G",null,"100","0.01","1000"]] 