explanation of Kirchhoff law and its application on solving electrical circuit

1. Kirchhoff’s laws
• Presented By:
• Vasudev Shrivastava
• P.G.T.(Physics)
• Jawahar Navodaya Vidyalaya
Nowgong
• District Chhatarpur (M.P.)

2. Kirchhoff’s lawsare fundamental to circuit theory. They quantify how current flows through a
circuit and how voltage varies around a loop in a circuit. Kirchhoff’s circuit laws were first
described in 1845 by the German physicist Gustav Kirchhoff.
Kirchhoff’s Junction Rule
Kirchhoff’s First Law
According to Kirchhoff’s Current Law, The total current entering a junction or a node
is equal to the charge leaving the node as no charge is lost. 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
The term Node refers to a junction or a connection
of two or more current-carrying routes like cables
and other components. Kirchhoff’s current law can
also be applied to analyze parallel circuits.

3. Kirchhoff’s first law are also known by several names as Kirchhoff’s
Current Law (KCL), Kirchhoff’s Junction Rule, Kirchhoff’s point rule,
Kirchhoff’s nodal rule. It is an application of the principle of conservation
of electric charge. The law states that at any circuit junction, the sum of
the currents flowing into and out of that junction are equal.
In simple terms, what KCL really says is that,
The sum of all currents entering a node is equal to the sum of all currents
leaving the node.
We perform analysis on all nodes based on the inflow and outflow of current.
Current directions at the node are based on presumed directions of the currents.
As long as the assumed directions of the currents are consistent from node to
node, the final result of the analysis will reflect the actual current directions in
the circuit.
Mathematically, Kirchhoff’s Current law is stated as follows
where n is the total number of branches carrying current towards or away from the node.
Solving Circuits Using KCL
Let us look at a few examples that demonstrate the law in practice and understand its importance
in determining the unknown parameters.
Example 1: Let us consider a network of assumed directions of the current as shown below.
Let us choose a sign convention such that currents entering the node are positive, and the currents
leaving the node are negative. With this convention, KCL applied at the node yields the equation:
i1(t)+i2(t)−i3(t)=0

4. or,
i1(t)+i2(t)=i3(t)
This amounts to the statement that the total current entering
the node is the same as the total current leaving the node.
Example 2: Using KCL to determine the value of the
unknown current i in the circuit below.
Assuming that the current entering the node is positive, the
sum of the currents is given by the equation,
4A−(−1A)−2A−i=0 ⇒4A+1A−2A=3A
Therefore, i = 3A, leaving the node(since our assumed
direction of i on the diagram is that it is leaving the node, and
our numerical value for i is positive).
Disadvantages of KCL
 KCL is valid only if the total electric charge is constant in
the circuit.
 KCL is suitable for high-frequency AC circuits.
Practice Questions
1. Calculate currents I_(1)and I_(2) in the given circuit?

5. Solution:
Let us identify the different nodes in
the circuit as A and B.
If the current entering the nodes are assumed positive, then 2 A
– I1 = 0. Therefore, I1 = 2 A.
Applying KCL at node B, we get the equation I1 + I2 = 0. Since
we know the value of I1, we can easily determine the value of
I2 as –2 A.
2. Calculate the currents I1 and I2, in the circuit below.
Solution:

6. Let us identify the two nodes in the circuit as
A and B.
Now applying KCL to node A, assuming the currents leaving the
nodes as positive, we get
Now applying KCL to node
B, we get
Substituting the value of I1 in the
above equation, we get
Solving, we get

7. Kirchhoff’s 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.
When you begin at any point of the loop and continue in the same direction, note the voltage drops in all
the direction either negative or positive and return to the same point. It is essential to maintain the
direction either counterclockwise or clockwise; else the final voltage value will not be equal to zero. The
voltage law can also be applied in analyzing circuits in series.
When either AC circuits or DC circuits are analysed based on Kirchhoff’s circuit laws, you need to be
clear with all the terminologies and definitions that describe the circuit components like paths, nodes,
meshes, and loops.
 Kirchhoff’s second law, also known as Kirchhoff’s voltage law (KVL) states
that the sum of all voltages around a closed loop in any circuit must be equal
to zero. This again is a consequence of charge conservation and also
conservation of energy.
Kirchhoff’s Voltage Law
Kirchhoff’s Second Law or the voltage 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. Hence,

8.  If a charge moves around a closed loop in a circuit, it must gain as
much energy as it loses.
 The above can be summarized as ” the gain in energy by the charge =
corresponding losses in energy through resistances
 Mathematically, the total voltage in a closed loop of a circuit is
expressed as ∑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, then 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.

9. Solving circuit using Kirchhoff’s Second Law

o 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.
o Defining our sign convention is very important
o Using Kirchhoff’s first law, at B and A we get, I1+I2=I3
o By making use of above convention and Kirchhoff’s Second Law
From Loop 1 we have:
10=R1∗I1+R3∗I3
=10I1+40I3 1=I1+4I3
From Loop 2 we have :
20=R2∗I2+R3∗I3

10. 20I2+40I3
1=I2+2I3
From Loop 3 we have :
10−20=10I1−20I2
1=−I1+2I2
 By making use of Kirchhoff’s First law I1+I2=I3
Equation reduces as follows (from Loop 1 ) :
1=5I1+4I2
Equation reduces as follows ( from Loop 2 ) :
1=2I1+3I2
 This results in the following Equation: I1=−13I2
 From last three equations we get, 1=13I2+2I2 I2=0.429A
I1=0.143AI3=0.286A
Kirchhoff’s Law Solved Example 02
If R1 = 2Ω, R2 = 4Ω, R3 = 6Ω, determine the electric current that
flows in the circuit below.
Solution:
1.
Following are the things that you should keep in mind while
approaching the problem:

11. You need to choose the direction of the current. In this problem, let
us choose the clockwise direction.
When the current flows across the resistor, there is a potential
decrease. Hence, V = IR is signed negative.
If the current moves from low to high then the source of emf (E) signed
positive because of the charging of energy at the emf source. Likewise, if
the current moves from high to low voltage (+ to -) then the source of
emf (E) signed negative because of the emptying of energy at the emf
source.
In this solution, the direction of the current is the same as the direction of
clockwise rotation.
– IR1 + E1 – IR2 – IR3 – E2 = 0
Substituting the values in the equation, we get
–2I + 10 – 4I – 6I – 5 = 0
-12I + 5 = 0 I = -5/-
12
I = 0.416 AThe electric current that flows in the circuit is 0.416 A. The
electric current is signed positive which means that the direction of the
electric current is the same as the direction of clockwise rotation. If the
electric current is negative then the direction of the current would be in
anti-clockwise direction.
Ncert Problem A battery of 10 V and negligible internal resistance
isconnected across the diagonally opposite corners of a cubical
networkconsisting of 12 resistors each of resistance 1 Ω (Fig. 3.23).
Determinethe equivalent resistance of the network and the current along
eachedge of the cube.

12. Example 3.7 Determine the current in each branch of the net work shown in Fig.
3.24.

15. Example 3.8 The four arms of a Wheatstone bridge (Fig. 3.26)
have the following resistances: AB = 100Ω, BC = 10Ω, CD = 5Ω,
and DA =60Ω.
]

17. Determine the current in each branch of the
network shown inFig. 3.30:
Advantages and Limitations of
Kirchhoff’s Law
The advantages of the laws are:
 It makes the calculation of unknown
voltages and currents easy
 The analysis and simplification of
complex closed-loop circuits becomes
manageable