Kirchhoffs Current law

Kirchhoff is also credited with developing the following equally important relationship between the currents of a network, called Kirchhoff's current law (KCL):

What is Kirchhoff's Current Law?

The algebraic sum of the currents entering and leaving a junction (or region) of a network is zero.
The law can also be stated in the following way:
The sum of the currents entering a junction (or region) of a network must equal the sum of the currents leaving the same junction (or region).
In equation form, the above statement can be written as follows:
$$\bbox[5px,border:1px solid grey]{ \sum I_i = \sum I_o } \tag{1}$$
with $I_i$ representing the current entering, or "in," and Io representing the current leaving, or "out."
Introducing Kirchhoff's current law
Fig. 1: Introducing Kirchhoff's current law.
In Fig. 1, for example, the shaded area can enclose an entire system or a complex network, or it can simply provide a connection point (junction) for the displayed currents. In each case, the current entering must equal that leaving, as required by Eq. (1):
$$\sum I_i = \sum I_o$$
$$ I_1 + I_4 = I_2 + I_3$$
$$ 4A + 8A = 2A + 10A$$
$$12A = 12A$$
In the next example, unknown currents can be determined by applying Kirchhoff's current law. Remember to place all current levels entering the junction to the left of the equals sign and the sum of all currents leaving the junction to the right of the equals sign.
In technology, the term node is commonly used to refer to a junction of two or more branches. Therefore, this term is used frequently in the analyses to follow.
Example 1: Determine currents $I_3$ and $I_4$ in Fig. 2 using Kirchhoff's current law.
Fig. 2: Two-node configuration for Example 1.
There are two junctions or nodes in Fig. 1. Node a has only one unknown, while node b has two unknowns. Since a single equation can be used to solve for only one unknown, we must apply Kirchhoff's current law to node a first.
At node a
$$\sum I_i = \sum I_o$$
$$ I_1 + I_2 = I_3$$
$$ 2A + 3A = I_3 = 5A$$
At node b, using the result just obtained,
$$\sum I_i = \sum I_o$$
$$ I_3 + I_5 = I_4$$
$$ 5A + 1A = I_4 = 6A$$

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