# 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.**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 I

_{IN}= Σ I

_{OUT}.

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 I_{T}leaving the junction is the algebraic sum of the two currents, I

_{1}and I

_{2}entering the same junction. That is I

_{T}= I

_{1}+ I

_{2}.

Note that we could also write this correctly as the algebraic sum of: I

_{T}- (I

_{1}+ I

_{2}) = 0.

So if I

_{1}equals 3 amperes and I

_{2}is equal to 2 amperes, then the total current, I

_{T}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 I

_{1}or I

_{2}. As I

_{1}= I

_{T}- I

_{2}= 5 - 2 = 3 amps, and I

_{2}= I

_{T}- I

_{1}= 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:To start, all the current, I

_{T}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 R

_{1}with the remainder continuing on through resistor R

_{2}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 R

_{1}

For current branch C to D through resistor R

_{2}

_{T}going into the junction at node B and two currents leaving the junction, I

_{1}and I

_{2}.

Since we now know from calculation that the currents leaving the junction at node B is I

_{1}equals 3 amps and I

_{2}equals 2 amps, the sum of the currents entering the junction at node B must equal 3 + 2 = 5 amps. Thus Σ

_{IN}= I

_{T}= 5 amperes.

In our example, we have two distinct junctions at node B and node E, thus we can confirm this value for I

_{T}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 Σ

_{IN}= I

_{T}= 5 amperes and therefore Kirchhoff’s current law holds true as this is the same value as the current leaving point A.

## Applying KCL to more complex circuits.

We can use Kirchhoff’s current law to find the currents flowing around more complex circuits. We hopefully know by now that the algebraic sum of all the currents at a node (junction point) is equal to zero and with this idea in mind, it is a simple case of determining the currents entering a node and those leaving the node. Consider the circuit below.### Kirchhoff’s Current Law Example No1

_{T}separates at node A flowing through resistors R

_{1}and R

_{2}, recombining at node C before separating again through resistors R

_{3}, R

_{4}and R

_{5}and finally recombining once again at node F.

But before we can calculate the individual currents flowing through each resistor branch, we must first calculate the circuits total current, I

_{T}. Ohms law tells us that I = V/R and as we know the value of V, 132 volts, we need to calculate the circuit resistances as follows.

### Circuit Resistance R_{AC}

### Circuit Resistance R_{CF}

_{T}is given as:

Giving us an equivalent circuit of:

### Kirchhoff’s Current Law Equivalent Circuit

_{AC}= 1Ω, R

_{CF}= 10Ω’s and I

_{T}= 12A.

Having established the equivalent parallel resistances and supply current, we can now calculate the individual branch currents and confirm using Kirchhoff’s junction rule as follows.

_{1}= 5A, I

_{2}= 7A, I

_{3}= 2A, I

_{4}= 6A, and I

_{5}= 4A.

We can confirm that Kirchoff’s current law holds true around the circuit by using node C as our reference point to calculate the currents entering and leaving the junction as:

_{1}+ I

_{2}- I

_{3}- I

_{4}- I

_{5}= 0 which equals 5 + 7 – 2 – 6 – 4 = 0.

So we can confirm by analysis that Kirchhoff’s current law (KCL) which states that the algebraic sum of the currents at a junction point in a circuit network is always zero is true and correct in this example.

### Kirchhoff’s Current Law Example No2

Find the currents flowing around the following circuit using Kirchhoff’s Current Law only._{T}is the total current flowing around the circuit driven by the 12V supply voltage. At point A, I

_{1}is equal to I

_{T}, thus there will be an I

_{1}*R voltage drop across resistor R

_{1}.

The circuit has 2 branches, 3 nodes (B, C and D) and 2 independent loops, thus the I*R voltage drops around the two loops will be:

- Loop ABC ⇒ 12 = 4I
_{1}+ 6I_{2} - Loop ABD ⇒ 12 = 4I
_{1}+ 12I_{3}

_{1}= I

_{2}+ I

_{3}, we can therefore substitute current I

_{1}for (I

_{2}+ I

_{3}) in both of the following loop equations and then simplify.

### Kirchhoff’s Loop Equations

Eq. No 1 : 12 = 10I

_{2}+ 4I_{3}Eq. No 2 : 12 = 4I

By multiplying the first equation (Loop ABC) by 4 and subtracting Loop ABD from Loop ABC, we can be reduced both equations to give us the values of I_{2}+ 16I_{3}_{2}and I

_{3}

Eq. No 1 : 12 = 10I

_{2}+ 4I_{3}( x4 ) ⇒ 48 = 40I_{2}+ 16I_{3}Eq. No 2 : 12 = 4I

_{2}+ 16I_{3}( x1 ) ⇒ 12 = 4I_{2}+ 16I_{3}Eq. No 1 – Eq. No 2 ⇒ 36 = 36I

_{2}+ 0Substitution of I

Now we can do the same procedure to find the value of I_{2}in terms of I_{3}gives us the value of I_{2}as 1.0 Amps_{3}by multiplying the first equation (Loop ABC) by 4 and the second equation (Loop ABD) by 10. Again by subtracting Loop ABC from Loop ABD, we can be reduced both equations to give us the values of I

_{2}and I

_{3}

Eq. No 1 : 12 = 10I

_{2}+ 4I_{3}( x4 ) ⇒ 48 = 40I_{2}+ 16I_{3}Eq. No 2 : 12 = 4I

_{2}+ 16I_{3}( x10 ) ⇒ 120 = 40I_{2}+ 160I_{3}Eq. No 2 – Eq. No 1 ⇒ 72 = 0 + 144I

_{3}Thus substitution of I

_{3}in terms of I_{2}gives us the value of I_{3}as 0.5 AmpsAs Kirchhoff’s junction rule states that : I

_{1}= I_{2}+ I_{3}The supply current flowing through resistor R

_{1}is given as : 1.0 + 0.5 = 1.5 AmpsThus I

We could have solved the circuit of example two simply and easily just using Ohm’s Law, but we have used _{1}= I_{T}= 1.5 Amps, I_{2}= 1.0 Amps and I_{3}= 0.5 Amps and from that information we could calculate the I*R voltage drops across the devices and at the various points (nodes) around the circuit.*Kirchhoff’s Current Law*here to show how it is possible to solve more complex circuits when we can not just simply apply Ohm’s Law.