For a given circuit, a connection of two or more elements is called a **NODE**. The particular circuit shown in figure 1 depicts an example of a node.

Figure.1: Circuit for Kirchhoff’s Current Law

We now present the Kirchhoff’s current law which is essentially the law of conservation of electric charge.

[stextbox id=’info’ caption=’Kirchhoff’s Current Law’]At any node of a circuit, the sum of the currents into the node is equal to the sum of the currents out of the node.[/stextbox]

Since charge must be conserved and does not accumulate at a node, the amount of current flowing out of a node equals the amount flowing in at any instant. In other words, an electrical node acts like a junction of water pipes where the amount of water going out equals the amount coming in.

Specifically for the portion of the network shown in figure 1, by applying KCL we obtain the equation

\[{{i}_{3}}+{{i}_{4}}={{i}_{1}}+{{i}_{2}}\]

An alternative, but equivalent, form of KCL can be obtained by considering currents directed into a node be positive in sense and currents directed out of a node to be negative in sense (or vice versa). Under this circumstance, the alternative form of KCL can be stated as follows:

[stextbox id=’info’ caption=’KCL in an Alternate Form ‘]At any node of a circuit, the currents algebraically sum to zero.[/stextbox]

$\sum{i}=0$

Applying this form of KCL to the node in figure 1 and considering currents directed into be positive in sense, we get

\[{{i}_{1}}+{{i}_{2}}-{{i}_{3}}-{{i}_{4}}=0\]

A close inspection of last two equations, however, reveals that they are the same.

- You May Also Read: Kirchhoff’s Voltage Law

**Kirchhoff’s current Law Solved Example**

Find I_{o } using KCL

Using KCL at node a,

$0.5{{i}_{0}}+3={{i}_{0}}$

${{i}_{0}}=6A$

**Application of KCL**

A very simple application of KCL is to combine current sources in parallel.

Using KCL, we can convert above figure into single current source form.

Current I_{1 } and I_{3 } are in the same direction as I_{T} total current, so we consider I_{1 } and I_{3 } as positive currents, while I_{2 } is in opposite direction of I_{1 } and I_{3} so will be considered as negative current. So, the total resultant current I_{T} will be,

${{i}_{T}}={{i}_{1}}-{{i}_{2}}+{{i}_{3}}$

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