Table of Contents
Introduction:
Kirchhoff’s Laws: these laws are use to find the current in different branches of an electrical network which may be easily solved by ohms law, Kirchhoff’s laws are two in number
 Kirchhoff’s first law or Kirchhoff’s current law “KCL”
 Kirchhoff’s loop law or Kirchhoff’s voltage law or Kirchhoff’s mesh law “KVL”.
Kirchhoff’s laws are applicable to both Ac and Dc.

Kirchhoff’s current Law
Kirchhoff’s current Law this law states in any electrical network that
“The algebraic sum of current meeting at a junction or node or point is zero”
Or
In other words “the sum of currents flowing towards the junction is equal to the sum of the currents flowing away from the junction”.
Let we consider the arrangement of 4wires connected together at point “0” as shown in fig below.
The arrows indicate the direction of flow of current. The currents I_{1} and I_{4} are coming towards the junction and current I_{2} and I_{3} are going out from junction we assume (suppose ) +ive sign for incoming current and –ive sign for outgoing current. Now according to the Kirchhoff’s Current Law “KCL”
∑I=0
Now from fig
I_{1}+I_{4}+(I_{2})+(I_{3})=0
I_{1}+I_{4}I_{2}I_{3}=0
I_{1}+I_{4}=I_{2}+I_{3}
Or incoming currents = outgoing currents
Problem1:
Determine the currents I3 and I4 for the giving circuit using Kirchhoff’s Current Law “KCL”.
Solution:
At point a by Kirchhoff’s
∑ I_{incoming }= ∑ I_{outgoing}
I_{1}+I_{2}=I_{3}
2+3=I_{3}
= I_{3} = 5
At point “b” by Kirchhoff’s law
∑ Iincoming = ∑Ioutgoing
I_{3} + I_{5} =I_{4}
5 +1=I_{4}
I_{4}=6_{amp}
Problem2:
Find I_{1}, I_{3}, I_{4}, and I_{5} in the network given
Solution:
At point a
I= I_{1}+I_{2}
5=I_{1}+4
54=I1
I_{1}=1_{amp}
As R1 and R3 are connected in series and we know that in series same current will pass through each resistor so
I_{3}=1_{amp}
And we know that R2 and R4 are connected in series so, current will be same therefore
I_{4}=4_{amp}
As I_{5} = I_{4} + I_{3}
I_{5} =4+1
I_{5}=5_{amp}
Problem3:
Find the magnitude and direction of I_{3} and I_{4}
I_{6} and I_{7} for the network using Kirchhoff’s Current Law KCL.
Solution:
Incoming current I_{1}= 10amp
Outgoing current I_{7}=10amp
At point b we see that
I_{2} = I_{4} +I_{5}
I_{2} = I_{4} +8
4= I_{4}
As at point d we see that
I_{7} = I_{5} + I_{6}
10 =8 +I_{6}
108 = I_{6}
I_{6} =2_{amp}
As at c point we see
I_{6 }+ I_{3} = I_{4}
2 +I_{3} = 4
I_{3} = 2_{amp}

Kirchhoff’s Voltage Law:
This law states that “the algebraic sum of all the voltage taken in a specified direction taken around the closed loop is zero”.
Mathematically
∑v=0
Closed loop:
A closed loop or a closed path is any continuous path that reaches a point in one direction and return to the same point from another direction without leaving the circuit.
Consider the circuit
Apply Kirchhoff’s Voltage Law KVL to the circuit we have
∑v=0
(xR_{1}) + (v_{1}) R_{2}(z)+ v_{2} –R_{3}(y)
xR_{1}v_{1}+R_{2}z+v_{2}+R_{3}y=0
For conformity the clock wise direction will be closed throughout all application of Kirchhoff’s Voltage Law KVL also anti clock wise direction can be used and both will gives us the same result.
Problem1:
Write the voltage equation for the given circuit
Solution:
v_{1}v_{2}+v=0
V=v_{1}+v_{2}
∑v_{rise} =∑v_{drop}
Problem2:
Find the voltage v_{1} for the network shown
Solution:
v_{1}v_{2}–∈_{2}+∈_{1}=0
Putting values
v_{1} 4.29+16=0
V_{1}= 4.2916
V_{1}=1613.2
V_{1}=2.8v_{out}