BASIC ELECTRICAL ENGINEERING

UNIT – I: D.C .Circuit

Objective:
To introduce the concepts of electrical circuits and its components
Learning outcomes:
To analyse and solve electrical circuits using network laws and
theorem

Learning Material
UNIT-1 DC CIRCUITS

1.1. Define following terms

(a) Current

Free electron Copper wire

Conventional current Electron current

Figure 1.1Concept of electric current

 Flow of electron in closed circuit is called current.

 Amount of charge passing through the conductor in unit time also called current.
 Unit of current is charge/second or Ampere (A).
Q
I
t
Where, I  Current
Q Charge
t  Time
(b) Potential or Voltage
 The capacity of a charged body to do work is called potential.
 Unit of potential is joule/coulomb or Volt (V).
W
V
Q
Where, V = Potential or Voltage
W = Workdone
(c) Potential difference
A B

+ 12 V +7V
Conventional current

Figure 1. 1Potential differences

 The difference of electrical potential between two charged bodies is called potential
difference.
 Unit of Potential Difference is Volt (V).
 If potential of body A is +12V and potential of body B is +7V then potential difference is
+5V.
i.e. (+12V) - (+7V) = +5V
UNIT-1 DC CIRCUITS

(d) Electro Motive Force (emf)

 The force is required to move electron from negative terminal to positive terminal of
electrical source in electrical circuit is called emf.
 Unit of emf is volt (V).
 Emf is denoted as ε.
(e) Energy
 Ability to do work is called energy.
 Unit of energy is Joule or Watt-sec or Kilowatt-hour (KWh).
 1KWh is equal to 1 Unit.
V 2t
W  P t  VIt  I2Rt 
R
Where, W=Energy
P =Power
t =Time

(f) Power
 Energy per unit in time is called power.
 Unit of Power is Joule/Second or Watt (W).
W
P
t
(g) Resistance
 Property of a material that opposes the flow of electron is called resistance.
 Unit of resistance is Ohm (Ω).
V
R=
I
Where, R  Resistance
(h) Conductance
 Property of a material that allows flow of electron.
 It is reciprocal of resistance.
 Unit of conductance is (Ω-1) or mho or Siemens(S).
1
G
R
Where, G  Conductance
(i) Resistivity or Specific Resistance
 Amount of resistance offered by 1m length of wire of 1m2 cross-sectional area.
 Resistivity is denoted as a ρ .
 Unit of Resistivity is Ohm-meter (Ω­m).
l
R
a
UNIT-1 DC CIRCUITS


l
Rρ 
a 
Ra
ρ 
l
Where, R  Resistance
ρ  Resistivity
l  Length of wire
a  Cross section area of wire

(j) Conductivity
 Ability of a material to allow flow of electron of a given material for 1 m length & 1
m2cross-sectional area is called conductivity. Unit of conductivity isΩ-1m-1 or Siemens m-
1.
1
σ
ρ
Where,σ  Conductivity
1.2. Explain types of electrical energysource
 Electrical source is an element which supplies energy to networks. There are two types
of electrical sources.
(a) Independent sources
Independent voltage source Independent current source

+ v(t)
V I i(t)
-

Figure 1. 2Independent voltage source Figure 1. 3Independent current source

 It is a two terminal element that provide a  It is two-terminal elements that provide a

specific voltage across its terminal. specific current across its terminal.
 The value of this voltage at any instant is  The value and direction of this current at
independent of value or direction of the any instant is independent of value or
current that flow through it. direction of the voltage that appears
across the terminal of source
UNIT-1 DC CIRCUITS

(b) Dependent sources

Voltage controlled voltage source (VCVS) Voltage controlled current source (VCCS)

+ + + Icd +
a c a c

Vab μV
+
Vcd
ab -
Vab gm Vab Vcd

b d b d
- - - -

Figure 1.5VCVS Figure 1.6VCCS

 Voltage controlled voltage source is four  Voltage controlled current source is four
terminal network components that terminal network components that
established a voltage Vcd between two- established a current icd in the branch of
point c and d. circuit.
Vcd  μVab icd  gmVab
 The voltage Vcd depends upon the control  icd depends only on the control voltage Vab
voltage Vab and μ is constant so it is and constant gm ,is called trans conductance
dimensionless. or mutual conductance.
 μ is known as a voltage gain.  Unit of transconductance is Ampere/Volt
or Siemens(S).

Current controlled voltage source (CCVS) Current controlled current source (CCCS)
iab icd
+ iab + + +
a c a c

ri + V
ab
- cd
β iab

b d b d
- - - -

Figure 1.7CCVS Figure 1.8CCCS

 Current controlled voltage source is four  Current controlled current source is four
terminal network components that terminal network components that
established a voltage Vcd between two- established a current Icd in the branch of
point c and d. circuit.
Vcd  riab icd  βiab
 Vcddepends on only on the control current  icd depends on only on the control current iab
iab and constant r and r is called trans and constantβ and β is called current gain.
resistance or mutual resistance. Current gain is constant.
 Unit of transresistance is Volt/Ampere  Current gain is dimensionless.
or Ohm (Ω).
UNIT-1 DC CIRCUITS

1.3. Explain source conversion

 A voltage source with a series resistor can be converted into an equivalent current
source with a parallel resistor. Conversely, a current source with a parallel resistor can
be converted into a voltage source with a series resistor.
 Open circuit voltages in both the circuits are equal and short circuit currents in both the
circuit are equal.Source transformation can be applied to dependent source as well.
R

I=V/R

Figure 1. 9Source conversion

Network simplification techniques

+ +

V1

V2

- -
(a)
+ +

V1

V2

- -
(V1 > V2)
(b)
UNIT-1 DC CIRCUITS

+ +

i1 i2 i1 + i2

-
(c)
+ +

i1 i2 i1 - i2

- -
(i1 > i2)
(d)
+ +

V1 = V2
V1 V2

- -
(e)
+ +

i1

i1 = i2
i2

- -
(f)
UNIT-1 DC CIRCUITS

Vs Vs

(g)
R + +

is
is

- -

(h)

Figure 1.10Rules under which source may be combined and separated

1.4. Explain ideal electrical circuit element.

 There are major three electrical circuit elements which are discussed below.
(a) Resistor
 Resistor is element which opposes the flow of current.

Figure 1.11Resistor Figure 1.12Conductor

 Resistance is property of material which opposes the flow current. It is measured in

Ohms (Ω).
 Value of resistance of conductor is
 Proportional to its length.
 Inversely proportional to the area of cross section.
 Depends on nature of material.
 Depends on temperature of conductor.
l
R
a
ρl
R
a
UNIT-1 DC CIRCUITS

(b) Inductor
 An inductor is element which store energy in form of magnetic field.
 The property of the coil of inducing emf due to the changing flux linked with it is known
as inductance of the coil.
 Inductance is denoted by L and it is measured in Henry (H).

 Value of inductance of coil is

1.13 Inductor

 Directly proportional to the square of number of turns.

 Directly proportional to the area of cross section.
 Inversely proportional to the length.
 Depends on absolute permeability of magnetic material.
F NI
Φ    NI  NIμ0 μr A
S S l l
μ0 μr A

 NIμ μ A
N  l0 r  N μ μ A
2

NΦ


Now, L    0 r
I I l
Where, L =Inductance of coil
N= Number of turns of coil
Φ = Flux link in coil
F = Magneto motive force(MMF)
I = Current in the coil
l = Mean length of coil
μ0 = Permiability of free space
μr = Relative permiability of magnetic material
A = Cross sectional area of magnetic material
(c) Capacitor
 Capacitor is an element which stored energy in form of charge.
 Capacitance is the capacity of capacitor to store electric charge.
 It is denoted by C and measured in Farad (F).

Figure 1.14Capacitor
 Value of capacitance is

 Directly proportional to the area of plate.

 Inversely proportional to distance between two plates.
UNIT-1 DC CIRCUITS

 Depends on absolute permittivity of medium between the plates.

UNIT-1 DC CIRCUITS

A
C
d
εA
C
d
ε0εr A
C
d
Where,C=Capacitance of capacitor
A =Cross sectional area of plates
d =Distance between two plates
ε = Abolute Permittivity
ε0 = Permittivity of free space
εr = Relative permittivity of dielectric material
1.5. Explain Ohm’s law and its limitations.
 Current flowing through the conductor is directly proportional to the potential difference
applied to the conductor, provided that no change in temperature.
Voltage (V)

Current (A)
Figure 1.15Change in current w.r.t change in voltage for conducting material
VI
V  IR
 Where R is constant which is called resistance of the conductor.
V
R 
I
 Limitations of Ohm’s Law:
 It cannot be applied to non-linear device e.g. Diode, Zener diode etc.
 It cannot be applied to non-metallic conductor e.g. Graphite, Conducting polymers
 It can only be applied in the constant temperature condition.
1.6. State and explain the Kirchhoff’s current and voltage laws
(a) Kirchhoff’s current law (KCL)
 Statement:
“Algebraic sum of all current meeting at a junction is zero”
 Let, Suppose
UNIT-1 DC CIRCUITS

 Branches are meeting at a junction ‘J’

 Incoming current are denoted with (+ve) sign
 Outgoing currents are denoted with (-ve) sign

A R2 I1 J I3 R4 B

I2
+
E1
- R3
R5
+
R1 E2 -

E D C
Figure 1.16Kirchhoff’s law diagram

 Then,
I  0
(  I1 ) (I2 ) (I3 )  0
I1  I2  I3  0

I1  I2  I3
Incoming current  Outgoing current
(b) Kirchhoff’s voltage law (KVL)
 Statement:
“Algebraic sum of all voltage drops and all emf sources in any closed path is zero”
 Let, Suppose
 Loop current in clockwise or anticlockwise direction
 Circuit current and loop current are in same direction than voltage drop is denoted
by (-ve) sign.
 Circuit current and loop current are in opposite direction than voltage drop is
denoted by (+ve) sign.
 Loop current move through (+ve) to (-ve) terminal of source than direction of emf
is (-ve).
 If Loop current move through (-ve) to (+ve) terminal of source than direction of
emf is (+ve).

+ R

I I
V= -IR V= +IR

E= -E1 E= +E1
UNIT-1 DC CIRCUITS

Figure 1.17Sign convention for Kirchhoff’s voltage law

UNIT-1 DC CIRCUITS

 IR   E  0
KVL to loop AJDEA
I1R2  I2R3  E2  I1R1  E1  0
KVL to loop JBCDJ
I3R4  I3R5  E2  I2R3  0
1.7. Explain series and parallel combination of resistor
Series combination of resistor Parallel combination of resistor
R1
I1

V1

I2 R2

- V2
V

Figure 1.18Series combination of resistors

Figure 1.19Parallel combinations of resistors

Here, I1  I2  I Here,V1  V2  V
As per KVL, As per KCL,
V  V1 V2 I  I1  I2
V  IR1  IR2 V V
I 
V  I(R1  R2 ) R1 R2
V  1 1 
 (R 1 R )2 IV  
I 
 R1 R2 
Req  R1  R2
I   1  1 
For n resistor are connected in series  
V  R1 R2 
Req  R 1 R 2 R  R
3
1  1 1 
 
Req R 1 R 2 
For n resistor are connected in Parallel
1  1 1 1 1
    ......... 
Req R1 R2 R3 Rn

 Value of equivalent resistance of series  Value of equivalent resistance of parallel

circuit is bigger than the biggest value of circuit is smaller than the smallest value of
individual resistance of circuit. individual resistance of circuit.
UNIT-1 DC CIRCUITS

1.8. Explain Voltage divider law and current divider Law.

Voltage Divider Law Current Divider Law
R1
I1
I1
R1
I2
R2 
V1
V1 V2 R2
I2

V2
-
V -
V

Figure 1.20Voltage divider circuit

Figure 1.21Current divider circuit

Here, I1  I2  I Here,V1  V2  V
As per KVL, As per KCL,

V  V1 V2 I  I1  I2
V  I1R1  I2R2 V V
I  1 2

V  IR  IR R1 R2
1 2
V V
V  I(R1  R2 ) I
V R1 R2

I  I1  I2   1 1 
(R  R )
1 2 I  V   
Now ,V1  I1R1  R 1 R 2 
V
V  R  1 1 
I
 
 

 

1 1
R R V R R
1
 R2   1 2 

V V VVVI R1R2 

1
 


1
RR 1 2  R  R 
 1 2   1 2 

Now,V2  I2R2
V  Now, I1 
V1
V  R2 R1
2

R1  R2  RR 

I R1 2
V  V  R2  I   1 R2 
R
  R 

1
2
R
1 2
 R1 
I I R 2 

UNIT-1 DC CIRCUITS

1   R 
UNIT-1 DC CIRCUITS


  1 2 
V
Now , I2  2
R2


 R R 
I R 1 2R
I2   1 2 

 RR2 
I I R 1 
2   R 
 1 2 
UNIT-1 DC CIRCUITS



 1.9. Derive the equation of delta to star and star to delta transformation


1 1
 
 

 
 

R12 R1 R31 R12 R1 R31
 
 

R2 R3 R2 R3
2 2

R23 R23

3 3
 
Figure 1.22Delta connected network Figure 1.23Star connected network

Resistance between terminal 1 & 2 Resistance between terminal 1 & 2
 R12 (R23  R31)  R1  R2
R (R  R 31) Resistance between terminal 2 & 3
 12 23
R12  R23  R31
 R2  R3
Resistance between terminal 2 & 3 Resistance between terminal 3 & 1
 R23 (R12  R31)  R3  R1
R (R  R31 )
 23 12
R12  R23  R31
Resistance between terminal 3 & 1
 R31 (R12 R23 )
R (R  R23)
 31 12
R12  R23  R31

Resistance between terminals 1 & 2 in delta equal to resistance between
terminals 1 & 2 in star
R  R  R12(R23  R31 ) (i)
1 2
RR R
12 23 31

Similarly,
RR R23(R12  R31 ) (ii)

2 3
RR R
12 23 31

R  R  R31(R12  R23 ) (iii)

3 1
RRR
12 23 31

(a) Delta to star conversion

Simplify i   ii  iii  on both the side of equations
R R R R-R-R R12(R23  R31 ) R23(R12  R31 ) R31(R12  R23 )
 + -
1 2 2 3 3 1 R  R R R  R R R  R R
12 23 31 12 23 31 12 23 31
UNIT-1 DC CIRCUITS

(R12R23  R12R31 ) (R23R12  R23R31 ) (R31R12  R31R23 )


+ -
R12  R23  R31 R12  R23  R R12  R23 R31
31

(R12R23  R12R31  R23R12  R23R31 - R31R12 - R31R23 )


(R12  R23  R31 )
2R  2R R
12 23

2
RR R
12 23 31
R12R23
R 
2
RR R
12 23 31

Similarly , R  R12R31

1
RR R
12 23 31
R23R31
R 
3
RR R
12 23 31

(b) Star to delta conversion

Simplify i ii   ii iii   iii i  on both the side of equation
(R1  R2 )(R2  R3 ) (R2  R3 )(R3  R1 ) (R3  R1 )(R1  R2 )
 (R  R )  R 23(RR12 RR31 )   R23(R1R2 RR
31 )
 R31(RR
12 RR  
23 )  R31(RR12 RR23 )
 R12(RR
23 
RR 
31 ) 
 R 12  R23  R31 
 12 23 31  12 23 31   12 23 31  12 23 31   12 23 31  12 23 31 

R1R 2 R R  R 22 R R2 3 R R 2 3R R 2 R1 2  R
1 3 3
R 3R1R  3R R 1
 R3 2 2 R R1 12
 R  R R31 R23R12  R23R31  R23R12  R23R31 R31R12  R31R23   R31R12  R31 23  12R23  R12 31  
 R12 23R 12 R RR R  R R R R R R R R R R R R

 12 23 31  12 23 31   12 23 31  12 23 31   12 23 31  12 23 31 
2 2 2
3R1R 2 3R 2R 
3
3R R 
3 1
R 
1
R 
2
R 3
 R 2R 2  R R 2R  R 2R R  R R R 2   R 2R R  R R 2R  R R R 2  R 2R 2   R 2R R  R 2R 2  R R 2R  R R R 2 
  23 12 R  R 12 R 23231 12 23 31    12 23 31 12R 23 R31  R12 223 31 23 31    12 23 31
12 23 31
R 31R
12 12 23 231
R  12 23 31 

 12 23 31  12 23 31  12 23 31 
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R
 23 12 12 23 31 12 23 31 12 23 31 12 23 31
R12 R 23R 31
12 23 31 12 22
3 31 23 31 12 23 31 12 31 12 23 31 12 23 31

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
(R R R  R R R  R R R  R R R  R R R  R R R  R R R  R R R  R R R ) (R R  R R  R R )
 12 23 31 12 23 31 12 23 31 12 23 31 12 23 31
R121223R 3123R 3112 2 23 31 12 23 31 12 23 31 23 12 23 31 12 31

2 2 2 2 2 2
RR R (R  R23 R 12 R 31R  R  R 23 R 31R ) (R R R R R R )
 12 23 31 12 2 12  23 31 23 31 23 12 1
2 2 31
R  R  R  R  R  R 

12 23 31 12 23 31

R R R (3R  3R  3R )  R R 2 2
R R 2 2 R R 2 2 
 R 23 2
 31   R2312R  R  2
 23R 31R  R  2 
 R 31R  R
12

2 
 
12 23 31 12
R R

12 23 31  12 23 31 12 23 31 12 23 31 

3R R R (R  R R) R 2R 2 R 2R 2 R 2R 2 
 R
12 23 31 12

23 2 31  23R12 R  R   2R3  R  R
2 31
 2  1R2  R  R 31
 2 
R R 
12 23 31  12 23 31 12 23 31 12 23 31 
 3R R  R 2  R 2  R 2


3 12 2 3 1

Now equation become

3R R  3R R  3R R  R 2  R 2  R 2  3R R  R 2  R 2  R 2
1 2 2 3 3 1 1 2 3 3 12 2 3 1

3R1R2  3R2R3  3R3R1  3R3R12

UNIT-1 DC CIRCUITS

R1R2
R R R
12 1 2
R3
Similarly
R2R3
R R R
UNIT-1 DC CIRCUITS

23 2 3
R1
R3R1
R R R
31 3 1
R2
UNIT-1 DC CIRCUITS

1.10. Explain Node analysis

R1 R3 R5
A B

+ +
V1 - R2 R4 -
V2

Figure 1.24Node analysis network

 Node: Node refers to any point on circuit where two or more circuit elements meet.
 Node analysis based on Kirchhoff’s current law states that algebraic summation of
currents meeting at junction is zero.
 Node C is taken as reference node in this network. If there are n nodes in any network,
the number of equation to be solved will be (n-1).
 Node A,B and C are shown in given network and their voltages areVA,VB andVC . Value of
node VC is zero because VC is reference node.
 Steps to follow in node analysis:
 Consider node in the network, assign current and voltage for eachbranch
and node respectively.
 Apply the KCL for each node and apply ohm’s law to branch current.
 Solve the equation for find the unknown node voltage.
 Using these voltages, find the required branch currents.
 Node A
R1 VA R3 VB R5
I1 I3

I2

V1 R2 R4 V2

VC

Figure 1.25Node analysis network for node A

Apply KCL at node A,
 I1    I2    I3   0
I1  I2  I3  0
VA - V1 VA VC VA - VB
  0
R1 R2 R3
 UNIT-1 DC CIRCUITS

1 1 1  1  V1
V   V   (i)
A  
R R R B  
R R
 1 2 3   3 1

 Node B
R1 R3 R5
VA I3 VB I5

I4

V1 R2 R4 V2

VC

Figure 1.26Node analysis network for node B

Apply the KCL at node B ,

 I3    I 4    I5   0
I3  I4  I5  0
VB - VA VB VC VB - V2
 R4  R5  0
R31 
V  V 1 1 1 V (ii)
  2

B  
A R R R R R
 3   3 4 5 5

From, equation (i) & (ii) 


 1  1 1 
1  
 V1 

 


V  R 


R R R R

 1 2 3   A    1 
3

 1 1 1  1  VB   V2 
 
    

  
R R R R R 
 3 3 4 5   5 

 One can easily find branch current of this network by solving equation (i) and (ii),if V1 ,
V2 and all resistance value are given.
1.11. Explain Mesh analysis
R1 R3 R5

V1 V2
UNIT-1 DC CIRCUITS

Figure 1.27Mesh analysis network

 Mesh: It is defined as a loop which does not contain any other loops within it.
UNIT-1 DC CIRCUITS

 The current in different meshes are assigned continues path that they do not split at a
junction into a branch currents.
 Basically, this analysis consists of writing mesh equation by Kirchhoff’s voltage law in
terms of unknown mesh current.

 Steps to be followed in mesh analysis:

 Identify the mesh, assign a direction to it and assign an unknown current in it.
 Assigned polarity for voltage across the branches.
 Apply the KVL around the mesh and use ohm’s law to express the branch voltage
in term of unknown mesh current and resistance.
 Solve the equations for unknown mesh current.
 Loop 1
R1 R3 R5

I1

V1 R2 R4 V2

I1 I2

Figure 1.28Mesh analysis network for loop-1

Now apply the KVL in loop  1,
- I1R1 -  I1  I2  R2  V1  0
- I1R1 - I1R2  I2R2 V1  0
-  R1  R2  I1  R2 I 2  V1 (i)

 Loop 2
R1 R3 R5

V1 V2

Figure 1.29Mesh analysis network for loop-2

Now Apply the KVL loop  2,

- I2 R3 -  I 2  I3  R4 -  I 2  I 1  R2  0
- I2R3 - I2R4 +I3R4  I2R2  I1R2  0
I1 R2 - I 2  R3  R4  R2  +I3 R4  0
UNIT-1 DC CIRCUITS

R2I1 -  R3  R4  R2  I 2 +R4 I 3  0 (ii)

UNIT-1 DC CIRCUITS

 Loop 3
R1 R3 R5

V1 V2

Figure 1.30Mesh analysis network for loop-3

Now Apply the KVL loop  3,
- I3R5 -V2 -  I3  I2  R4  0
- I3R5 -V2 -I3R4  I2R4  0
I2R4 - I3  R5  R4   V2

R4 I2 -  R5  R4  I3  V2 (iii)

From equation (i),(ii) &(iii)

 -  R1  R2  R2 0  I1   V1 
R R  R  R  R
    
I  0
  R  R   2 
4 2 4
 2 3 R  
0
4 I 
5 4  V 
   3  2 
 -  R1  R2  R2 0 
Δ R R  R  R  R
 2 3 4 2 4 
0 R   R5  R4  
 4 
 V1 R2 0 
 
Δ  0 R  R  R  R 
1  3 
R5  R4  
4 2 4
V R4
2

 
 -  R1  R2  V1 0 
Δ  R 0 R
  R5  R  
4
2
 0
2
V2
4
 
 -  R1  R2  R2 V1 
 
Δ  R R  R  R  0 
3
 2 3 4 2 
0 R4 V2
 
Now ,
Δ Δ Δ
I  1,I  2,I  3
1
Δ 2 Δ 3 Δ
UNIT-1 DC CIRCUITS

1.12. Explain Superposition theorem

 The superposition theorem states that in any linear network containing two or more
sources, the current in any element is equal to the algebraic sum of the current caused by
individual sources acting alone, while the other sources are inoperative.
UNIT-1 DC CIRCUITS

 According to the application of the superposition theorem. It may be noted that each
independent source is considered at a time while all other sources are turned off or killed.
To kill a voltage source means the voltage source is replaced by its internal resistance
whereas to kill a current source means to replace the current source by its internal
resistance.
 To consider the effects of each source independently requires that sources be removed
and replaced without affecting the final result. To remove a voltage source when applying
this theorem, the difference in potential between the terminals of the voltage source must
be set to zero (short circuit) removing a current source requires that its terminals be
opened (open circuit).
 Any internal resistance or conductance associated with the displaced sources is not
eliminated but must still be considered.
 The total current through any portion of the network is equal to the algebraic sum of the
currents produced independently by each source.
 That is, for a two-source network, if the current produced by one source is in one
direction, while that produced by the other is in the opposite direction through the same
resistor, the resulting current is the difference of the two and has the direction of the
larger.
 If the individual currents are in the same direction, the resulting current is the sum of two
in the direction of either current. This rule holds true for the voltage across a portion of a
network as determined by polarities, and it can be extended to networks with any
number of sources.
 The superposition principle is not applicable to power effects since the power loss in a
resistor varies as the square (nonlinear) of the current or voltage.
 Steps to be followed to apply the superposition theorem:
 Select any one energy source.
 Replace all the other energy sources by their internal series resistances for
voltage sources. Their internal shunt resistances for current sources.
 With only one energy source calculate the voltage drops or branch currents
paying attention to the voltage polarities and current directions.
 Repeat steps 1, 2 and 3 for each source individually.
 Add algebraically the voltage drops or branch currents obtained due to the
individual source to obtain the combined effect of all the sources.
 Example network:

R1 A R2

R3 -
V1 V2

Figure 1.31Superposition theorem network

UNIT-1 DC CIRCUITS

Step-1

R1 A R2

V1 +

I1 R3 I2

Figure 1.32Superposition theorem network for step-1

Now apply Mesh analysis in loop  1,
- I1R1 - I1R3  I2R3 - I1r V1  0
Now apply Mesh analysis in loop  2,
- I2R2 - I2R3  I1R3  0
Now, current flow from R3 branch is a lg ebric sum of I1 and I2

Step-2

R1 A R2

+
I3 R3 I4
-
r V2

B
Figure 1.33Superposition theorem network for step-2

Now apply Mesh analysis in loop  1,

- I3R1 - I3R3  I4R3 - I3r  0
Now apply Mesh analysis in loop  2,
- I4R2 -V2 - I4R3  I3R3  0
Now, current flow from R3 branch is a lg ebric sum of I3 and I4
Finally, current flow from R3 is alg ebric sum of step1 and step-2

1.13. Explain Thevenin’s theorem

 Thevenin theorem is an analytical method used to change a complex circuit into a
simple equivalent circuit consisting of a single resistance in series with a sourcevoltage.
 Thevenin’s can calculate the currents and voltages at any point in a circuit.
 Thevenin’s Theorem states that “Any linear circuit containing several voltages and
resistances can be replaced by just one single voltage in series with a single resistance
UNIT-1 DC CIRCUITS

Connected across the load“.

 In other words, it is possible to simplify any electrical circuit, no matter how complex, to
an equivalent two-terminal circuit with just a single constant voltage source in series with
a resistance (or impedance) connected to a load as shown below.
 Thevenin’s Theorem is especially useful in the circuit analysis of power or battery
systems and other interconnected resistive circuits where it will have an effect on the
adjoining part of the circuit.
 Thevenin’s equivalent circuit

A Linear Network
containing
Several emf’s and
Resistance

Figure 1.34Thevenin’s equivalent circuit

 As far as the load resistor RL is concerned, any complex “one-port” network consisting of
multiple resistive circuit elements and energy sources can be replaced by one single
equivalent resistance Rth and one single equivalent voltage Eth.
 Rth is the thevenin resistance value looking back into the circuit and Eth is the Thevenin’s
voltage (open circuit voltage) at the terminals.
 Steps to be followed to apply the Thevenin’s theorem:
 Remove the load resistor Rth or component concerned.
 Find Rth by shorting all voltage sources or by open circuiting all the current
sources.
 Find Eth by the usual circuit analysis methods.
 Find the current flowing through the load resistor Rth.
 Example network:
R1 R2

R3 RL

B
Figure 1.35Thevenin’s theorem network
UNIT-1 DC CIRCUITS

Step-1
R1 R2 A
+
V1 +
-

R3 Eth
I1 I2
r

-
B
Figure 1.36Thevenin’s theorem network (step-1)

Now apply Mesh analysis in loop  1,

- I1R1 - I1R3  I2R3 - I1r V1  0
Now apply Mesh analysis in loop  2,
- I2R2 - Eth - I2R3  I1R3  0
Loop - 2 is open that ' s way I2  0,
So, Eth  I1R3
Eth  Thevenin equivalent voltage
Rth  Theveninequivalent Re sis tan ce
RL  Load Re sis tance
Step-2

R1 R2 A

Rth  r+ R  R + R 
1 3 2

R3  r+ R  R  
Rth 1 3
Rth   +R
r r+ R1   R3  2

  

B
Figure 1.37Thevenin’s theorem network (step-2)

Step-3

Rth

IL
Eth
+ IL 
RL Rth R
- L
Eth

Figure 1.38Thevenin’s theorem network (step-3)

UNIT-1 DC CIRCUITS
UNIT-1 DC CIRCUITS

1.14. Explain Norton’s theorem

 Norton’s theorem is an analytical method used to change a complex circuit into a simple
equivalent circuit consisting of a single resistance in parallel with a current source.
 Norton’s Theorem states that “Any linear circuit containing several energy sources and
resistances can be replaced by a single Constant Current generator in parallel with a
Single Resistor“.
 As far as the load resistance, RL is concerned this single resistance, RN is the value of the
resistance looking back into the network with all the current sources open circuited and
IN is the short circuit current at the output terminals as shown below.
 Norton’s equivalent circuit
A

IN

Figure 1.39Norton’s theorem equivalent circuit

 The value of this “constant current” is one which would flow if the two output terminals
where shorted together while the Norton’s resistance would be measured looking back
into the terminals.
 The basic procedure for solving a circuit using Norton’s Theorem is as follows:
 Remove the load resistor RL or component concerned.
 Find RN by shorting all voltage sources or by open circuiting all the current
sources.
 Find IN by placing a shorting link on the output terminals A and B.
 Find the current flowing through the load resistor RL.
 Example network:

RL

B
Figure 1.40Norton’s theorem network
UNIT-1 DC CIRCUITS

Step-1
R1 R2

V1 +

I1 R3 I2 IN

B
Figure 1.41 Norton’s theorem network (step-1)
Now apply Mesh analysis in loop  1,
- I1R1 - I1R3  I2R3 - I1r V1  0
Now apply Mesh analysis in loop  2,
- I2R2 - I2R3  I1R3  0
Here I2  IN
IN  Norton' s equivalent current
RN  Norton' s equivalent Re sis tance
RL  Load Re sis tance
Step-2

R1 R2 A

RN  r+ R  R + R 
1 3 2
 r+ R  R  
R3 RN   1 3
+R
r+ R1   R3 
RN 2

r   

B
Figure 1.42 Norton’s theorem network (step-2)

Step-3

IL

RN
IN RN RL IL  I N
R R
N L

Figure 1.43 Norton’s theorem network (step-3)

UNIT-1 DC CIRCUITS

1.15. Time domain analysis of first order RC circuit

Charging of Capacitor Discharging of Capacitor
R R
+ - - +
VR VR
+ + + -
V C V C
- VC - V
- C
+

Figure 1.44Charging of capacitor Figure 1.45Discharging of capacitor

Apply KVL in circuit , Apply KVL in circuit ,

V -VR -Vc  0 0  VR  Vc
V  VR  Vc 0  iR  Vc
dq
V  iR  Vc 0  R  Vc
dq dt
V RV d CVc 
dt
c 0 R V
d CVc  dt c
V RV dVc
dt c
0  RC V
dV dt c
V  RC c  V
dt c dVc
Vc -RC
dV dt
V -V  RC c
1 -1

c
dt dV 
1 1 V c  RC dt
 dV 
c  RCdt c
V -V c -t
Multiply min us sign both the side log V    K (i)
c
RC
-1 -1
 dV 
c  RC dt When, t  0, Vc  V
V -V c log V   K (ii)
-t
logV -V    K (i) Solve equation (i) and (ii)
c
RC -t
log V    log V 
When, t  0, Vc  0 c
RC
-t
log V   K (ii) logV - logV  
Solve equation (i) and -t
(ii) c
RC
log V -V    logV 
log  
Vc  -t
c

-tRC   RC
V
logV -V - logV    Vc  RC-t
c Ve
RC  
-t
log 
V -V c  -t
  Vc  VeRC
 V  RC-t
 V -V  
c
 Ve
RC

 
 Vc  -t
1 -   e RC
 V 
UNIT-1 DC CIRCUITS

-t

Vc  V(1- e ) RC

dq dq
Also, i  Also,i 
dt dt
d(CVc ) d(CVc )
i i
dt dt
d -t dVc
i  C (V (1 - e RC )) iC
dt -t dt -t
d d
i  VC (1-e RC) i  C (VeRC )
dt dt -t
-1
  1  RC-t  i  CV
i  VC  0-  - e  eRC
RC
  RC   V -t

VC RC-t i  - e RC
i e
RC R
-t
V -t i  -Ime
i  eRC RC

R
-t

i  ime RC

V λ

0.632 V

vc
vc
0.37

t λ t
Figure 1.46Charging voltage of capacitor Figure 1.48Dicharging voltage of capacitor

O
λ t
I

-0.37 Im

ic

0.37

-Im

λ
Figure 1.47Charging current of capacitor Figure 1.49Dicharging current of capacitor

1.16. Time domain analysis of first order RL circuit

UNIT-1 DC CIRCUITS

Charging of Inductor Discharging of Inductor

R
+ -

VR + VR
+
V - V 
VL L VL L

Figure 1.50Charging of inductor Figure 1.51Discharging of inductor

From KVL,
From KVL,
di di
V - iR - L  0 -iR - L  0
dt dt

di di
-iR L
V - iR  L dt dt

di -R
di dt   di
 
i L
V - iR L
1 1 1 -R
 di  dt   di  dt
 V - iR L
i L
 -R 
-R -R logi   t K (i)

V - iRdi  L dt
  L 
 

V 
logV -iR 
 -R 
t K When, t  0, i 
 L  (i)
R


V
When, t  0, i  0 log K (ii)
 R 
log V   K (ii)  

Solve (i) and (ii)
Solve equation (i) and (ii)  -R  V 
 -R  logi   t  log
logV -iR  t  logV   L 


     
L  -R  logi - log V   - R  

t
logV -iR  - logV    t  R   L 
     
 L 
 V - iR   -R  log  i  
 -R 

 

   L 
t t
log  V   L 
 V - iRV -Rt   
R  -R 
 L
 e   
 V 
 UNIT-1 DC CIRCUITS

i  t


V 
    e  L
 R -R 
L t
 R 
1-  i  e


 i V  -Rt
L
 V  e  

  -R t  R
UNIT-1 DC CIRCUITS

i 
V 
1- e L V


    i  e λt

R  R
1- e-Rt 
i  I m L
 
 
i  Im
1- e   λt
UNIT-1 DC CIRCUITS



λ

  -R 
t
0.632 Im i  Im 1 - e L   
 
  
V
 i e λt
iL R
iL
0.37 Im

t λ t

Figure 1.52Charging current of inductor Figure 1.53Dicharging current of inductor

 Assignment Cum Tutorial Questions

PART – A: Objective Questions

(Blooms Levels: BL1 – Remembering; BL2 - Understanding; BL3 – Applying)

SHORT ANSWER TYPE QUESTIONS:

1. Define (i) Resistor (ii) Inductor (iii) capacitor
2. Explain Active and Passive elements.
3. Difference between ideal source and practical source.
5. Classify all the different types of voltage and current sources
6. Differentiate between independent and dependent sources?
7. Explain voltage- current relationship for passive elements?
8. Explain source transformation with suitable examples.
13. State Thevenins theorem
14. State Norton’s theorem for D.C circuits
15. State and explain Superposition theorem & its limitations
17. List the advantages of Thevinins and Nortons theorem
18. How to convert Thevinins equivalent circuit to Nortons equivalent circuit
 PART – B: Subjective Questions

(Blooms Levels: BL2 - Understanding; BL3 – Applying; BL4 – Analyzing)

ESSAY ANSWER TYPE QUESTIONS:

1. Breifly explain about different types of sources
2. Two coils A and B have resistance of 12Ω and 6Ω and inductances of 0.02H and 0.03H
respectively. These are connected in parallel and a voltage of 200v at 50hz is applied to their
combination. Find (a) Current in each coil (b) total current (c) power factor of the current (d)
power consumed by each coil and total power
3. By using KVL and KCL find the currents in each elements
4. By using nodal analysis find the current through 3 ohm resistor
5. Find the equivalent resistance across the terminal A- B as shown in figure
6. Determine V1 and V2 in the circuit shown in below figure
7. Determine the current supplied by each battery in the circuit shown in figure by using
Kirchhoff’s laws.
8. State and explain Norton’s theorem for A.C Excitation?
9. Stat and explain the superposition theorem for A.C Excitation? (b) State and explain
reciprocity theorem for D.C Excitation?
10. Explain the Phenomenon of resonance circuits? Derive the formula for the resonant
frequency of the series resonant circuit
11. Compare series resonance and parallel resonance circuit. An RLC circuit consists of R
=1kΩ, L = 100mH C= 10µF. If a voltage of 100v is applied across the combination,
determine resonant frequency, Q factor and Bandwidth?
12. (a) For a series resonance circuit obtain the expression for bandwidth in terms of
resonance frequency and Q factor. (b) For an RL series circuit having variable resistance and
fixed reactance draw the current, impedance and admittance Loci
13. Find the voltage across the 2Ω resistor by using Superposition theorem
14. Derive the expression for Q factor bandwidth in parallel RLC circuit
15. Obtain the expression for frequency at which maximum voltage occurs across the
capacitor in series in resonance circuit in terms of Q factor and resonance frequency
16. Derive the transient equation for RL series circuit
17. Derive the transient equation for RL series circuit