Kirchhoff’s circuit law is the basic law in the circuit analysis field. In 1845 a german physicist Gustav Robert Kirchhoff was the first to develop formulas for governing the behavior of an electrical circuit. He gave two laws :

- The first law is known as Kirchhoff’s current law which is also known as first law or point law.
- The second law is known as Kirchhoff’s voltage law which is also known as the second law or mesh law.

Table of Contents

## Kirchhoff’s Current Law ( KCL )

Kirchhoff’s current law states that the algebraic sum of current at a node or a junction of a circuit at any time is always zero. This means the current entering the node is equal to the current leaving the node.

So, if you are taking current entering the node as positive then you have to take current leaving the node as negative. From the above figure,

I_{1 }+ I_{4} = I_{2} + I_{3 }+ I_{5, } ………. Eq 1.

current I_{1 }& I_{4 }are entering the node which is taken as positive and current I_{2}, I_{3 }& I_{5 }are leaving the node which is taken as negative. You can write Eq 1. as

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

## Kirchhoff’s Voltage Law ( KVL )

Kirchhoff’s voltage law states that at any time the algebraic sum of voltage within a loop has to be zero. In other words, the total voltage in a loop is equal to the total voltage drop across each element.

So in loop ‘abca’, the sum of voltage is

E_{1} – V_{1 } – V_{2 }= 0 …………. Eq 2.

OR

E_{1} = V_{1} + V_{2}

This law is applicable regardless of the direction of the loop. The direction of the loop does not have to be the same as the direction of the current. But you may look for a sign convention.

Now, let’s see a question.

Q. Find the current flowing across the resistor R_{3} in the fig. given below.

By KCL :

At node C ,

I_{1} + I_{2} = I_{3}

So current across R_{1} = I_{1}

Current across R_{2} = I_{2}

Current across R_{3} = I_{1} + I_{2}

Now by KVL :

In loop 1 :

E_{1} – E_{2} = I_{1}R_{1} + I_{2}R_{2}

10 – 5 = 10I_{1} +10I_{2}

5 = 10I_{1} + 10I_{2} ………………………… eq. 3

In loop 2 :

E_{2} = I_{2}R_{2} + I_{3}

5 = 10I_{2} + 5( I_{1} + I_{2 })

5 = 5I_{1} + 15I_{2} ………………. eq. 4

Now by solving above eq. 3 & 4

Current across resistor R_{2}

I_{2 }= ¼

Current Across Resistor R_{1}

I_{1 }= ½

so the current across resistor R_{3}

I_{3 }= I_{1 }+ I_{2}

I_{3} = ¼ + ½

= ¾