KIRCHHOFF'S LAWS

Kirchhoff's laws are very useful in solving circuits which cannot be easily solved by Ohm's law.

First Law: "The algebraic sum of all the currents meeting at any junction point in an electric circuit is zero".
This law relates to the currents at the junction points of a circuit.
Consider four wires meeting at a point O and carrying currents I1, I2, I3 and I4 as shown in fig.

Sign conventions of currents:
●Currents flowing towards junction point O taken as positive(+ve) sign, and
●Currents flowing away from junction point O taken as negative(-ve) sign.

According to Kirchhoff's first law,
Algebraic sum of currents meet at a junction point '0' = 0  i.e.  ∑ I = 0

And according to sign conventions of currents I1 and I2 = +ve,

I3 and I4 = -ve

Therefore Kirchhoff's first law is
(I1) + (I2) + (- I3) + (- I4) = 0
or
(I1) + (I2) - (I3) - (I4) = 0 
or
(I1) + (I2) = (I3) + (I4)
i.e Sum of incoming currents = Sum of outgoing currents.

Therfore Kirchhoff's first law also stated that,
The algebraic sum of incoming currents is equal to the algebraic sum of outgoing currents at any junction point in an electric circuit.

Kirchhoff's Second Law: "In any closed circuit the algebraic sum of products of currents and resistances (voltage drops) plus the algebraic sum of all the e.m.fs. in that circuit is zero".

In any closed circuit,
Algebraic sum of e.m.f's. + Algebraic sum of voltage drops = 0  
or      ∑ E  +  ∑ IR  =  0

Sign conventions of e. m. f's and voltage drops:

● Sign of e. m. f's : A rise in potential should be considered positive(+ve) while a fall in potential should be considered negative(-ve). Thus, if we go from the -ve terminal of a battery to the +ve terminal, there is a rise in potential and it should be taken as positive(+ve). On the other hand, if we go from the +ve terminal of a battery to the -ve terminal, there is a fall of potential and it should be taken as negative(-ve). It should be noted that sign of e.m.f. is independent of the direction of current.

● Sign of voltage drops : There is a voltage drop in resistance due to the flow of current through it. If we go with the current, the voltage drop should be taken as negative(-ve). On the other hand, if we go against or opposite direction of the current flow, the voltage drop should be taken positive(+ve). It should be noted here that sign of voltage drop depends upon the direction of current and is independent of e.m.f. in the circuit.