Kirchhoff’s Current Law

Kirchhoff’s Current Law (KCL) is Kirchhoff’s first law that deals with the conservation of charge entering and leaving a junction.

To determine the amount or magnitude of the electrical current flowing around an electrical or electronic circuit, we need to use certain laws or rules that allows us to write down these currents in the form of an equation. The network equations used are those according to Kirchhoff’s laws, and as we are dealing with circuit currents, we will be looking at Kirchhoff’s current law, (KCL).

Gustav Kirchhoff’s Current Law is one of the fundamental laws used for circuit analysis. His current law states that for a parallel path the total current entering a circuits junction is exactly equal to the total current leaving the same junction. This is because it has no other place to go as no charge is lost.

In other words the algebraic sum of ALL the currents entering and leaving a junction must be equal to zero as: Σ IIN = Σ IOUT.

This idea by Kirchhoff is commonly known as the Conservation of Charge, as the current is conserved around the junction with no loss of current. Lets look at a simple example of Kirchhoff’s current law (KCL) when applied to a single junction.

A Single Junction

Here in this simple single junction example, the current ITleaving the junction is the algebraic sum of the two currents, I1and I2 entering the same junction. That is IT = I1 + I2.

Note that we could also write this correctly as the algebraic sum of: IT - (I1 + I2) = 0.

So if I1 equals 3 amperes and I2 is equal to 2 amperes, then the total current, IT leaving the junction will be 3 + 2 = 5 amperes, and we can use this basic law for any number of junctions or nodes as the sum of the currents both entering and leaving will be the same.

Also, if we reversed the directions of the currents, the resulting equations would still hold true for I1 or I2. As I1 = IT - I2 = 5 - 2 = 3 amps, and I2 = IT - I1 = 5 - 3 = 2 amps. Thus we can think of the currents entering the junction as being positive (+), while the ones leaving the junction as being negative (-).

Then we can see that the mathematical sum of the currents either entering or leaving the junction and in whatever direction will always be equal to zero, and this forms the basis of Kirchhoff’s Junction Rule, more commonly known as Kirchhoff’s Current Law, or (KCL).

Resistors in Parallel

Let’s look how we could apply Kirchhoff’s current law to resistors in parallel, whether the resistances in those branches are equal or unequal. Consider the following circuit diagram:

 

In this simple parallel resistor example there are two distinct junctions for current. Junction one occurs at node B, and junction two occurs at node E. Thus we can use Kirchhoff’s Junction Rule for the electrical currents at both of these two distinct junctions, for those currents entering the junction and for those currents flowing leaving the junction.

To start, all the current, IT leaves the 24 volt supply and arrives at point A and from there it enters node B. Node B is a junction as the current can now split into two distinct directions, with some of the current flowing downwards and through resistor R1 with the remainder continuing on through resistor R2 via node C. Note that the currents flowing into and out of a node point are commonly called branch currents.

We can use Ohm’s Law to determine the individual branch currents through each resistor as: I = V/R, thus:

For current branch B to E through resistor R1

For current branch C to D through resistor R2

 

From above we know that Kirchhoff’s current law states that the sum of the currents entering a junction must equal the sum of the currents leaving the junction, and in our simple example above, there is one current, IT going into the junction at node B and two currents leaving the junction, I1 and I2.

Since we now know from calculation that the currents leaving the junction at node B is I1equals 3 amps and I2 equals 2 amps, the sum of the currents entering the junction at node B must equal 3 + 2 = 5 amps. Thus ΣIN = IT = 5 amperes.

In our example, we have two distinct junctions at node B and node E, thus we can confirm this value for IT as the two currents recombine again at node E. So, for Kirchhoff’s junction rule to hold true, the sum of the currents into point F must equal the sum of the currents flowing out of the junction at node E.

As the two currents entering junction E are 3 amps and 2 amps respectively, the sum of the currents entering point F is therefore: 3 + 2 = 5 amperes. Thus ΣIN = IT = 5 amperes and therefore Kirchhoff’s current law holds true as this is the same value as the current leaving point A.