Kirsof Currents Law

Kirsof Currents Law (KCL) is the first law of Kirsof, which deals with the protection of the load that enters and exits a node.

To determine the amount or size of the electric current flowing around an electrical or electronic circuit, we need to use certain laws or rules that allow us to write these currents in the form of an equation.The network equations used are according to Kirchhoff's laws, and we will use Kirchhoff's current law (KCL) when dealing with circuit currents.

Gustav Kirchhoff's Current Actis one of the basic laws used for circuit analysis.Current law states that the total current entering a circuit connection for a parallel path is exactly equal to the total current separated from the same connection.This is because there is no loss of current, so there is nowhere else to go.

In other words, the algebraic sum of ALL currents entering and exiting a connection should be equal to zero as follows: Ε I IN = Ε I OUT 

Kirchhoff's idea is widely known as The Protection of the Load, since the current is protected without loss of current around the connection.Let's look at a simple example of Kirchhoff's current law (KCL) when applied to a single node.

One Node

Kirchhoff's Currents Law

Here in this simple single connection example, the current separated from the connection I T is the algebraic sum of the I 1 and I 2 current entering the same connection.That is, I T = I 1 + I 2 .

Note that we can also spell this correctly as algebraic sum: I T – (I 1 + I 2 ) = 0 .

So if I 1 equals 3 ampere and I 2 equals 2 amps, then the total current will be 3 + 2 = 5 amps separated from the I T joint, and we can use this basic law for any number of connections or nodes as follows: the sum of the currents in and out will be the same.

Also, if we reverse the directions of the currents, the resulting equations will still apply to I 1 or I 2 .I 1 = I T – I 2 = 5 – 2 = 3 amps and I 2 = I T – I 1 = 5 – 3 = 2 amps.Thus, we can consider the currents entering the connection as positive (+) and the allocated currents as negative (-).

Then we can see that the mathematical sum of currents entering or exiting the connection, regardless of which direction, will always equal zero, and this forms the basis of Kirchhoff's Node Rule, or (KCL), more commonly known as Kirchhoff's Current Act.

Parallel Resistors

Now let's look at how we can apply Kirsof's current law to parallel-linked resistors,the resistances in these branches may not be equal or equal.Consider the following circuit diagram:

Kirchhoff's Currents Law

In this simple example of parallel resistance, there are two different connections for the current.Firstthe node occurs on node B and the second connection occurs on node E. Thus, we can use the Kirsophobia Currents Act (KCL) for electrical currents in these two different nodes, currents entering the connection and currents coming out of the connection.

To begin with, the entire current, I T, leaves the 24-volt source and reaches point A, from there it enters node B. Node B is a connection as part of the current flows downwards, while the current can now be divided in two different directions, and is called branch currents for their currents, which continue with the R1 resistance and the remaining R2 resistance.

We can use the Ohm Law to determine individual branch currents through each resistance: I = V/R, so:

For current branch from B to E via R1 resistance

Kirchhoff's Currents Law

For current branch C to D through R2 resistance

Kirchhoff's Currents Law

From above, the law of kirsophag currents, the sum of the currents entering a node,we know that it indicates that the currents coming out of the node should be equal to the sum, and in our simple example above, there is current I T entering node B andThere are two currents separated from the node, I 1 and I 2.

Now for currents separated from node B: 3 amps equals I1 and I2 equals 2 amps, Bthe sum of currents entering from the node must be equal to 3 + 2 = 5 amps.Thus, Ε IN = I T = 5 amps.

In our example, we have two different connections on node B and node E, so we can verify this value for I T as the two currents recombin on node E. Thus, for Kirchhoff's connection rule to be valid, the sum of the currents must be equal to the sum of the currents flowing from the connection on node E.

Since the two currents entering the E connection are 3 amps and 2 amps respectively, the sum of the currents entering point F is therefore: 3 + 2 = 5 amps.Thus, Ε IN = I T = 5 amps and therefore Kirsof'sthe law of currents applies, because this is the same value as the current output point A.

Applying KCL to More Complex Circuits

We can use Kirchhoff's current law to find currents flowing around more complex circuits.We said that the algebraic sum of all currents in a node (junction point) is equal to zero. A simple way to determine currents entering and leaving a node.Consider the following circuit:

Kirchhoff's Currents Law Question Example 1

Kirchhoff's Currents Law

In this example, the current has four different node points.

But before calculating the individual currents flowing from each branch of resistance, we must first calculate the total current of the circuit, I T.Ohm law tells us that it is I = V/R, and since we know the value of V, 132 volts, we need to calculate circuit resistors as follows.

Circuit Resistance RAC

Kirchhoff's Currents Law

Thus, the equivalent circuit resistance between nodes A and C is calculated as 1 Ohm.

Circuit Resistance R CF

Kirchhoff's Currents Law

Thus, the equivalent circuit resistance between nodes C and F is calculated as 10 Ohm.Then the total circuit current, I T is given as follows:

Kirchhoff's Currents Law

Kirchhoff's Equivalent Circuit of the Currents Act

Kirchhoff's Currents Law

Therefore, V = 132V, R AC = 1Ω, R CF = 10Ωs and I T = 12A.

After determining the equivalent parallel resistors and feed current, we can now calculate individual branch currents and verify Kirchhoff's connection rule using the following.

Kirchhoff's Currents Law

Thus, I 1 = 5A, I 2 = 7A, I 3 = 2A, I 4 = 6A and I 5 = 4A.

To calculate the currents entering and exiting the connection as follows, we use node C as our reference point towe can confirm that the law of currents is correct around the circuit:

Kirchhoff's Currents Law

In addition, while the currents entering the unit are positive, we can double-check whether Kirchhoff's Current Law is correct because the node exits are negative, so the algebraic total is equal to: I 1 + I 2 – I 3 – I 4 – I 5 = 0, 5 + 7 – 2 – 6 – 4 = 0.

Therefore, we can confirm by analyzing that Kirchhoff's current law (KCL), which states that the algebraic sum of currents on a port in a circuit network is always zero, is correct in this example.

Kirchhoff's Currents Law Question Example 2

Find the currents flowing in the following circuit only using Kirchhoff's Current Law.

Kirchhoff's Currents Law

I T is the total current flowing around the circuit driven by the 12V supply voltage.

There are 2 arms, 3 nodes (B, C and D) and 2 independent loops in the circuit, so the I*R voltage drops around the two loops will be as follows:

  • Loop ABC ⇒ 12 = 4I 1 + 6I 2
  • Loop US ⇒ 12 = 4I 1 + 12I 3

Since Kirchhoff's law of currents indicates that in node B, there is I 1 = I 2 + I 3, therefore, we can replace the I 1 current (I 2 + I 3) in both of the following cycle equations and then simplify it.

Kirchhoff's Cycle Equations

Kirchhoff's Currents Law

Now we have two simultaneous equations about currents flowing around the circuit.

Equation 1: 12 = 10I 2 + 4I 3

Equation 2: 12 = 4I 2 + 16I 3

By multiplying the first equation (ABC Cycle) by 4 and removing Loop US from Loop ABC, we can simplify both equations to give us the values I 2 and I 3.

Equation 1: 12 = 10I 2 + 4I 3 ( x4) ⇒ 48 = 40I 2 + 16I 3

Equation 2: 12 = 4I 2 + 16I 3 ( x1) ⇒ 12 = 4I 2 + 16I 3

Equation 1 –Equation 2 ⇒ 36 = 36I 2 + 0

From these equations, we calculate I2 as 1 amp.

Now we can do the same to find the value of I 3 by multiplying the first equation (Loop ABC) by 4and the second equation (Loop US) by 10. Equations that give us I 2 and I 3 values:

Equation1: 12 = 10I 2 + 4I 3 ( x4) ⇒ 48 = 40I 2 + 16I 3

Equation 2: 12 = 4I 2 + 16I 3 ( x10 ) ⇒ 120 = 40I 2 + 160I 3

Equation 2 –Equation 1 ⇒ 72 = 0 + 144I 3

We calculate I3 as 0.5 amps from these equations.

As Kirchhoff points out in the node rule: I 1 = I 2 + I 3

The feed current passing through the R 1 resistance is given as follows: 1.0 + 0.5 = 1.5ampere

So I 1 = I T = 1.5 amps , I 2 = 1.0amperage and I 3 = 0.5From amps and this information we can calculate I*R voltage drops between devices and at various points (nodes) around the circuit.

We could solve the circuit in the second example simply and easily using the Ohm Act – but here we showed how it is possible to solve complex circuits using only Kirchhoff's Current Law.