- Circuit analysis overview
- Kirchhoff's current law
- Kirchhoff's voltage law
- Kirchhoff's laws
- Labeling voltages
- Application of the fundamental laws (setup)
- Application of the fundamental laws (solve)
- Application of the fundamental laws
- Node voltage method (steps 1 to 4)
- Node voltage method (step 5)
- Node voltage method
- Mesh current method (steps 1 to 3)
- Mesh current method (step 4)
- Mesh current method
- Loop current method
- Number of required equations
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.
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,
- Label the voltages and currents using the sign convention for passive components.
- Select the independent variable, either or 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 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:
- nodes, labeled , , and .
- meshes (inner loops).
- source voltage, , and element voltages, and .
- source current, , and element currents, , and .
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 and .
(There is an obvious opportunity here to simplify the two parallel resistors, with . 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 or current ? One good way to make this choice is to compare the number of unknown voltages to unknown currents. There are unknown voltages, and unknown currents. If we select voltage as the independent variable, we will have equations with voltage terms as opposed to current terms. 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 . Why did I make these particular choices? I picked the two most interesting features of the circuit. Node 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 . 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.
The signs for and are because we encounter their 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
We will get our second equation by writing Kirchhoff's Current Law at node . 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 , set them equal to the sum of currents flowing out.
Earlier in Step 2. we decided to use and as the independent variables, so we use Ohm's Law to express the unknown currents in terms of voltage and resistance.
After a little rearrangement, we have our system of two equations in two unknowns,
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 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 and , and unknown currents , , and .
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 or to produce the simplest equations. Choose the variable with the fewest unknowns.
- Write equations using KCL, KVL, or both. Make sure every element is represented in at least one equation.
- Solve the system of equations for the independent variables (in this case, and ).
- 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 ?(11 votes)
- 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.)(15 votes)
- 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?(6 votes)
- 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.(10 votes)
- Why isn't there a voltage over the 5 ohm resistor?(3 votes)
- The voltage across the 5 ohm resistor is v2.
v2 is the voltage between nodes b and c, so this is the voltage appearing across both resistors and the current source.(4 votes)
- excuse-me, at the lesson " step3- KCL at node B" <<-- how and where to get that i1 + is = i2 + i3;
i2, i3 and iS have the same node. I still don't understand it. Please help.(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.(4 votes)
- 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.(3 votes)
- Can you explain more about the current divider rule(2 votes)
- "Current divider" is the nickname given to a circuit made of parallel resistors. In a parallel connection, the main current splits and flows through the individual resistors. You are interested in the share of current that goes through one of the resistors.
Suppose you have two resistors in parallel, R1 and R2. Draw them vertically. Assign a current called I_total that comes in from the top left and leaves from the bottom left. Figure out the current in each resistor.
If you know I_total, R1, and R2, you can figure out the voltage, V.
V = I_total x R_equivalent = I_total x (R1 || R2) "||" means "in parallel"
V = I_total ( R1 R2 / (R1 + R2)
That's the expression for V. Now figure out the current in an individual resistor. That's just Ohm's Law for the resistor. For R1 you get,
I_R1 = V / R1 Ohm's Law
Now replace V with the expression derived above and you should get the Current Divider Formula,
I_R1 = I_total (( R2 / R1 + R2))
I've never used a current divider, but the "voltage divider" circuit comes up thousands of times. I think the only role for current dividers is to be a homework problem.
Voltage divider references:
Video: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-resistor-circuits/v/ee-voltage-divider(1 vote)
- please can i get help with how the line that follows "and crank the algebra" in the solution came about.
ho did we get -540/30? please help.(1 vote)
- The first term on the right side of the equation as a 30 in the denominator. The second term is -18. -18 is equal to -540/30. Now both terms have a 30 in the denominator.(2 votes)
- 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)
- The thing that's causing confusion is the definition of node c. Node c is the entire bottom horizontal wire in the circuit. So the two segments you identified (between voltage source and 6ohm, and the segment between two resistors) are both included as part of node c. Review the definition of node and "distributed node" in this article https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/circuit-elements/a/ee-circuit-terminology. When you account for the currents flowing into node c you will draw arrows on the vertical wire segments feeding into node c. You won't be drawing current arrows on the horizontal bits and pieces of node c. It's all one single junction.(2 votes)
- 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.
- In step 1, you don't treat any part of the circuit as a parallel circuit. So there is no difference in how you analyse the parallel circuit in comparison to the series circuit? Is this a standard in Kirchoff's Laws? Is the reason for this because, in a round about way, we have already resolved the currents (or voltages) accordingly so that Ohm's law, as it pertains specifically to parallel and series circuits, doesn't apply anymore? For example (still in step1) i(3) = i(1) - i(2). Does that make sense?(1 vote)
- In the simple example circuit there are two resistors in parallel, so there is an opportunity to simplify the circuit by combining them. However, in this lesson I wanted to leave those two resistors in place so I could focus on the method of solving the circuit. I didn't want the circuit to become too simple before getting to the point of the lesson. This wasn't meant to suggest a standard method of applying Kirchhoff's Laws.(2 votes)