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):
with $I_i$ representing the current entering, or "in," and Io representing the
current leaving, or "out."
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):
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.
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:$$\bbox[5px,border:1px solid grey]{ \sum I_i = \sum I_o } \tag{1}$$
Fig. 1: Introducing Kirchhoff's current law.
$$\sum I_i = \sum I_o$$
$$ I_1 + I_4 = I_2 + I_3$$
$$ 4A + 8A = 2A + 10A$$
$$12A = 12A$$
Example 1: Determine currents $I_3$ and $I_4$ in Fig. 2 using Kirchhoff's current law.
Solution:
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
At node b, using the result just obtained,
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
$$\sum I_i = \sum I_o$$
$$ I_1 + I_2 = I_3$$
$$ 2A + 3A = I_3 = 5A$$
$$\sum I_i = \sum I_o$$
$$ I_3 + I_5 = I_4$$
$$ 5A + 1A = I_4 = 6A$$
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