Encyclopedia of Electrical Engineering

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 law can also be stated in the following way:
In equation form, the above statement can be written as follows:

with Ii representing the current entering, or "in," and Io representing the
current leaving, or "out."
**Fig.no.1: ** Introducing Kirchhoff's current law.
In Fig. No. 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. No. 2 using Kirchhoff's current law.
**Fig.No.2: **Two-node configuration for Example 1.
**Solution:**There are two junctions or nodes in Fig. no.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, $$\sum I_i = \sum I_o$$ $$ I_3 + I_5 = I_4$$ $$ 5A + 1A = I_4 = 6A$$
##### Kirchhoffs Current law Related Questions

$$\bbox[5px,border:1px solid red] {\color{blue}{ \sum I_i = \sum I_o }}$$ | Eq.(1) |

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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