Kirchhoff's Current Law (KCL)
Why Do You Need To Know About KCL?
Using KCL To Write Equations For Circuits
Problems
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Kirchhoff's Current Law - Introduction

Kirchhoff's Current Law - KCL - is one of two fundamental laws in electrical engineering, the other being Kirchhoff's Voltage Law (KVL).

• KCL is a fundamental law, as fundamental as Conservation of Mass in mechanics, for example, because KCL is really conservation of charge.
• KVL and KCL are the starting point for analysis of any circuit.
• KCL and KVL always hold and are usually the most useful piece of information you will have about a circuit after the circuit itself.
•  People and computer programs both use KVL and KCL for circuit analysis. Spice (in all its incarnations) starts with KCL.

Goals For This Lesson

What should you be able to do after this lesson?  Here's the basic objective.

Given an electrical circuit:
Be able to write KCL at every node in the circuit.
Be able to solve the KCL equations, especially for simple circuits.
These goals are very important.  If you can't write KCL equations and solve them, you may well be lost when you take a course in electronics in a few years.  It will be much harder to learn that later, so be sure to learn it well now.

Kirchhoff's Current Law

At this point, you have learned the fundamentals of charge and current.  There is one important law, Kirchhoff's Current Law that you will need to learn.  It is not as complex as it might seem.  All you really need to know is that charge is conserved, so KCL is really based on one simple fact.

• Charge can neither be created nor destroyed.
From that basic fact we can get to Kirchhoff's Current Law.  Despite that simplicity, it is a fundamental, widely used law, that you need to know to go very far in electrical engineering.

Let's examine a circuit simulation.  It's shown below.  Charge (current) is flowing through the circuit.  The simulation shows some charge - the large red blob - flowing through a battery (where it picks up energy, but that's another story.  Click here for that lesson.)  That charge flows through Element #1 in the simulation.  After the charge flows through Element #1 it splits.  Some of the charge goes through Element #2, and some goes through Element #3.  (Notice that it does not split equally!  Sometimes it does.  Sometimes it doesn't.)  When, in the course of its flow through the circuit, there is no possibility of splitting, all of the charge entering a node will flow through the next element.  (That element is said to be in series.  Element #3 and Element #4 are in series because all of the current going through #3 goes through #4.  Elements #1 and #2 are not in series.)

There is one node in the simulation where charge flowing through two elements comes together and "reunites" and flows back into the battery.

Note that this simulation emphasizes the conservation of charge.  When charge flows through Element #1 when it gets to the end of Element #1 it splits into two.  However, what arrives at that node is what leaves that node, so the amount of charge that enters the node - the big red blob - equals the amount of charge that leave that node - the sum of the charge on the medium sized red blob and the charge on the small red blob.

Problem

1.  In this circuit, charge flows from the battery, through Element #1 to the node.  Willy Nilly observes that 35 coulombs flows through Element #1 in 20 seconds, and that, in that same time, 17 coulombs flows through Element #2.  How much charge flows through Element #3 in that time?

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

2.  How much charge flows through Element #4 in that time?
Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

KCL

Charge usually flows through some sort of metallic wire, flowing through the atomic lattice.  Although it is physically unlike water flowing in a pipe, that analogy is sometimes drawn. Like water confined to the interior of a pipe, charge is confined to flow within a wire, and it doesn't leave the surface of the wire. You may want to think in those terms as you interpret current flow in the sketches and diagrams that follow.  We will be developing rules for current flow in circuits in this section.  You will need to know about that in order to be able to analyze larger circuits with lots of elements.

In practice current flows in wires and often splits between two or more devices.  We need to consider what happens in networks of conductors in which current can split.  Single wires carrying current aren't the most important case we can look at, and you need to learn about Kirchhoff's Current Law which describes those situations where we have large networks of interconnected elements carrying current.  Those kinds of circuits will have many connection points (called nodes) where current can split into smaller currents.  Shown below is part of a circuit.  Current (I) comes in from the left and splits into two parts, I1 and I2.  There is one simple relationship between these two currents and the current, I, flowing in from the left below.

Here, a red dot has been placed over both of the nodes in the picture.

Focus attention on a very short time, DT.  Assume all currents constant during DT.

• A current, I, flows into the top node, and I1 and I2 are flowing out of the node.  No charge accumulates!
• During time DT, the total amount of charge that flows into the node is zero so:
• IDT - I1DT - I2DT  = 0
• And during DT,
• IDT is the charge flowing in.
• I1DT is the charge flowing through the left resistor.
• I2DT is the charge flowing through the right resistor.
• So, we have - for the period of time DT, the total amount of charge that flows into the node is zero so:
• IDT - I1DT - I2DT  = 0
• Cancelling the DT's everywhere, we get:
• I - I1 - I2  = 0
• Which can be rephrased as:
• The sum of the currents flowing into the node is zero.
• or
• I  =  I1 + I2
• which says that "The current entering the node equals the current leaving the node."
We need to be more precise in this.
• When we have the expression:
• I - I1 - I2  = 0
• Or when we think "The sum of the currents flowing into the node is zero."
• We interpret I as a current entering and - I1 and  - I2 also as currents entering.  Note the negative signs!
• Since I1 is leaving the node, then we can think of  - I1 as the value of the current entering.
• We do the same for I2 and - I2.
Again, we need to be more precise when we express things the other way.
• When we have the expression:
• I = I1 + I2  = 0
• Or when we think "The sum of the currents flowing into the node equals the sum of the currents leaving the node."

• We interpret I as a current entering and I1 and  I2 as currents leaving.
Either of the above formulations is Kirchhoff's Current Law, otherwise known as KCL.  If you understand that "The sum of the currents entering a node is zero.", then you know KCL.  With that, it's time for you to answer a few questions.

Problems

3.  In this circuit - which you saw above - determine the current I2, in terms of trhe other two currents.  You will need to write KCL at the node marked with a red dot.  Notice that we have defined current symbols and polarities for all the currents involved.

4.  Now, determine a value for the current, I2, when you have numerical values for the other currents.
• I1 = .75 A
• I3= - .45 A (nobody said the value had to be positive!)
Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

5.  Here is another circuit.

You need to determine a value for the current, I4, given the following numerical values for some other currents.  First, you'll need to get an algebraic expression for I4.  Click on the corrrect expression.

• Now, determine the numerical value for I4 when I3 is 0.45A.
Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

6.  Here's a problem for you.  In 10 seconds, an observer - Willy Nilly - notices that 35 coulombs of charge leaves node "n" in this circuit, heading for node "x".  (Vn is the voltage at node "n", etc.)  In the same ten seconds, 22 coulombs of charge leaves node "n" heading for node "z".  Determine the current, Iy.

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.