Kirchhoff’s Second Law states that the “net electromotive force around a closed circuit loop is equal to the sum of potential drops around the loop”. It is termed as Kirchhoff’s Loop Rule which is an outcome of an electrostatic field which is conservative.

\(\sum V=0\) The below figure illustrates that the total voltage around a closed loop must be zero.

This law manages the voltage drops at different branches in an electrical circuit. Consider one point on a closed loop in an electrical circuit. If somebody goes to another point in a similar ring, he or she will find that the potential at that second aspect might be not quite the same as the first point.

On the off chance that he or she keeps on setting off to some unique point on the loop and he or she may locate some extraordinary potential in that new area. If he or she goes on further along that closed loop, eventually he or she achieves the underlying point from where the voyage was begun.

That implies, he or she returns to a similar potential point in the wake of the intersection through various voltage levels. It can be then again said that the gain in electrical energy by the charge is equal to corresponding losses in energy through resistances.

# Examples of Kirchoff’s Second Law

The first and foremost step is to draw a closed loop to a circuit. Once done with it draw the direction of the flow of current.

- By using Kirchhoff’s First Law

At B and A

\(I_{1}+I_{2}=I_{3}\)

- By making use of above convention and Kirchoff’s Second Law

From Loop 1 we have :

\(10=R_{1}*I_{1}+R_{3}*I_{3}\)

\(=10I_{1}+40I_{3}\)

\(1=I_{1}+4I_{3}\)

From Loop 2 we have :

\(20=R_{2}*I_{2}+R_{3}*I_{3}\)

\(20I_{2}+40I_{3}\)

\(1=I_{2}+2I_{3}\)

From Loop 3 we have :

\(10-20=10I_{1}-20I_{2}\)

\(1=-I_{1}+2I_{2}\)

By making use of Kirchhoff’s First law \(I_{1}+I_{2}=I_{3}\)

Equation reduces as follows (from Loop 1 ) :

\(1=5I_{1}+4I_{2}\)

Equation reduces as follows ( from Loop 2 ) :

\(1=2I_{1}+3I_{2}\)

This results in the following Equation:

\(I_{1}=-\frac{1}{3}I_{2}\)

From last three equations we get,

\(1=\frac{1}{3}I_{2}+2I_{2}\)

\(I_{2}=0.429A\)

\(I_{1}=0.143A\)

\(I_{3}=0.286A\)

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