Kirchhoff’s Circuit Law

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 :

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

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.

Kirchhoff's circuit law - KCL

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,

I1 + I4 = I2 + I3 + I5,  ………. Eq 1.

current I1 & Iare entering the node which is taken as positive and current I2, I3 & Iare leaving the node which is taken as negative. You can write Eq 1. as

I1 + I4 – I2 – I3 – I5 = 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.

Kirchhoff's circuit law - KVL

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

E1 – V1  – V2 = 0  …………. Eq 2.


E1 = V1 + V2

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 R3 in the fig. given below.

Question on Kirchhoff's circuit law

By KCL :

At node C ,

I1 + I2 = I3

So current across R1 = I1

Current across R2 = I2

Current across R3 = I1 + I2

Now by KVL :

In loop 1 :

E1 – E2 = I1R1 + I2R2

10 – 5 = 10I1 +10I2

5 = 10I1 + 10I2 ………………………… eq. 3

In loop 2 :

E2 = I2R2 + I3

5 = 10I2 + 5( I1 + I2 )

5 = 5I1 + 15I2 ………………. eq. 4

Now by solving above eq. 3 & 4

Current across resistor R2

I2 = ¼

Current Across Resistor R1

I1 = ½

so the current across resistor R3

I3 = I1 + I2

I3 = ¼ + ½

= ¾

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