1 Mohammad Waqas
(19CH18)
Presentation Topic: Kirchhoff's Laws
Subject: Basic ELECTRICAL TECHNOLOGY
Date: 24/7/2020

Kirchhoff's Laws
 Kirchhoff's laws quantify how current flows through a circuit and how
voltage varies around a loop in a circuit
 There are two laws
 Kirchhoff’s Current Law (KCL) or First Law
 Kirchhoff’s Voltage Law (KVL) or Second Law
Kirchhoff’s Current Law (KCL) or First Law
 The total current entering a junction or a node is equal to the charge
leaving the node as no charge is lost
2

 Put differently, the algebraic sum of every current
entering and leaving the node has to be null. This
property of Kirchhoff law is commonly called as
Conservation of charge wherein, I(exit) + I(enter) =
0.
 In the above figure, the currents I1, I2
and I3 entering the node is considered
positive, likewise, the currents I4 and I5
exiting the nodes is considered negative
in values. This can be expressed in the
form of an equation:
 I1 + I2 + I3 – I4 – I5 = 0
Figure
3

 Node
 It is the point in a circuit at which at least two elements
(active or passive) are joined.
 Junction
 It is the point in a circuit at which at least three
elements (active or passive) are joined.
 Note:-A junction must be a node but a node may or
may not be a junction
4
Loop
 A loop current is defined as a constant current that
flows around a closed path or loop. (A closed path is a
path through the network that ends where it starts.)

Kirchhoff’s Voltage Law (KVL) or Second
Law
 According to Kirchhoff’s Voltage Law,
 The voltage around a loop equals to the sum of
every voltage drop in the same loop for any closed
network and also equals to zero.
 Put differently, the algebraic sum of every voltage in
the loop has to be equal to zero and this property of
Kirchhoff’s law is called as conservation of energy.
5
Figure

6
 Kirchhoff's Law Rules

 Example Of KCL
 Calculate the value of current I in the section of networks shown in
figure
At junction A : 𝐼2 + 8 = 15 ; 𝐼2 =7A
At junction D : 𝐼3 + 5 = 8 ; 𝐼3 = 3A
At junction B : 𝐼2 + 3 =𝐼4; 𝐼4 =10A
At junction C : 𝐼1= 𝐼3 + 𝐼4 ; 𝐼1 =13 A
 Solution
 Figure
7

 EXAMPLE OF KVL
 Find out V1 and V2 using KVL
 Solution
 -20 + 𝑉1+ 𝑉2 = 0 ….. Equation 1
 𝑉1= 2 і
 𝑉2 = 3 i
 -20 + 2 i + 3 i = 0
 5 i = 20
 I = 4A
 𝑉1= 2 i = 2(4) = 8V
 𝑉2 = 3 i = 3(4) = 12 V
Put In Equation 1
 FIGURE
8

9