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 I₅. The current I₁, I₂, and I₃ are entering the node N, while current I₄ and I₅ 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 I₁, I₂, and I₃ will be positive, while current I₄ and I₅ will be negative.

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

Therefore, I₁ + I₂ + I₃ – I₄ – I₅ = 0

∴ I₁ + I₂ + I₃ = I₄ + I₅ ——-(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, Q₁ + Q₂ + Q₃ = Q₄ + Q₅ ———-(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, I₂ = 2A, I₄ = -1A and I₅ = -4A then find current I₆

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

Solution:

Applying KCL at node B,

I₃ + I₆ = I1 and I₁ = 2A (Given)

∴ I₆ = 2 – I₃ ——— (3)

Similarly, applying KCL at node C,

I₂ + I₅ = I₃ and I₅ = -4 A (Given)

∴ I₃ = I₂ – 4 ———-(4)

Similarly, applying KCL at node A,

I₁ + I₄ = I₂ and I₄ = -1 A and I₁ = 2A (Given)

∴ I₂ = 2 -1 = 1A

From equation 4, I₃ = 1 – 4 = -3A

And by putting the value of I₃ in equation 3,

I₆ = 2 – (-3) = 5A

Therefore, for the given circuit, current I₆ = 5A

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

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