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 I1, I2, I3, I4, 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: