Kirchhoff’s Current Law (KCL) is also known as Kirchhoff’s first law or Kirchhoff’s junction law.

**Statement : **“Algebraic sum of branch currents at node is zero at all instance of time.”

Or

“At any node (junction) in a network, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.”

**Fig.1 Electrical Network with 5 branches**

To understand the statement, let’s consider one example. As shown in Fig. 1, the network has one node N and 5 branches. The current in five branches is_{ }I_{1, }I_{2, }I_{3, }I_{4,} and I_{5}. The current I_{1}, I_{2}, and I_{3} are entering the node N, while current I_{4} and I_{5 }are leaving the node.

**Sign Convention**

If current is entering the node, then consider it as positive current. Similarly, if current is leaving the node then it can be considered as negative current. The same is shown in Fig.2.

**Fig.2 Sign Convention for Kirchhoff’s Current Law (KCL**)

Therefore, for the network shown in Fig. 1, current I_{1}, I_{2}, and I_{3} will be positive, while current I_{4} and I_{5} will be negative.

And according the Kirchhoff’s Current Law, the algebraic sum of all these current is zero.

Therefore, I_{1} + I_{2 }+ I_{3} – I_{4} – I_{5 }= 0

**∴** I_{1} + I_{2} + I_{3} = I_{4} + I_{5} ——-(1)

It shows that, the sum of currents entering the node is equal to the sum of currents leaving the node.

**KCL is a law of Conservation of Charge**

The current is the rate at which the charge is flowing.

**∴** Current (I) = Q/t

Therefore, the equation 1 can be written as

And further after the simplification, Q_{1} + Q_{2} + Q_{3} = Q_{4 }+ Q_{5} ———-(2)

Equation 2 shows that, the charge which is entering the node is equal to the charge which is leaving the node.

Therefore, Kirchhoff’s Current Law (KCL) is the law of conservation of charge.

**Example**

For the given circuit if, I_{2} = 2A, I_{4} = -1A and I_{5} = -4A then find current I_{6}

**Fig. 3 Kirchhoff’s Current Law (KCL) Example **

**Solution:**

Applying KCL at node B,

I_{3 }+ I_{6} = I1 and I_{1} = 2A (Given)

**∴** I_{6} = 2 – I_{3} ——— (3)

Similarly, applying KCL at node C,

I_{2} + I_{5} = I_{3} and I_{5} = -4 A (Given)

**∴** I_{3} = I_{2} – 4 ———-(4)

Similarly, applying KCL at node A,

I_{1} + I_{4} = I_{2} and I_{4} = -1 A and I_{1} = 2A (Given)

**∴** **I _{2} = 2 -1 = 1A**

From equation 4, **I _{3} = 1 – 4 = -3A**

And by putting the value of I_{3 }in equation 3,

**I _{6} = 2 – (-3) = 5A**

Therefore, for the given circuit, current I_{6} = 5A

For more information on Kirchhoff’s Current Law (KCL) check this video: