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1Concept of electric current





Flow of electron in closed circuit is called current
...

Unit of current is charge/second or Ampere (A)
...

Unit of potential is joule/coulomb or Volt (V)
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1Potential differences





The difference of electrical potential between two charged bodies is called potential
difference
...

If potential of body A is +12V and potential of body B is +7V then potential difference is
+5V
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e
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D
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Circuits
(d)



(e)




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
...

Emf is denoted as ε
...

Unit of energy is Joule or Watt-sec or Kilowatt-hour (KWh)
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W  P  t  VIt  I 2Rt 

V 2t
R

Where , W =Energy
P =Power
t =Time

(f)



Power
Energy per unit in time is called power
...


P
(g)



(h)




W
t

Resistance
Property of a material that opposes the flow of electron is called resistance
...

V
R=
I
Where , R  Resistance
Conductance
Property of a material that allows flow of electron
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Unit of conductance is (Ω-1) or mho or Siemens(S)
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Resistivity is denoted as a ρ
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R

l
a

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Unit of conductivity isΩ-1m-1 or Siemens m1
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There are two types
of electrical sources
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2Independent voltage source




Figure 1
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The value of this voltage at any instant is
independent of value or direction of the
current that flow through it
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The value and direction of this current at
any instant is independent of value or
direction of the voltage that appears
across the terminal of source

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Vcd  μVab

circuit
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μ is known as a voltage gain
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Unit of transconductance is Ampere/Volt
or Siemens(S)
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8CCCS

Current controlled voltage source is four
terminal network components that
established a voltage Vcd between two-



point c and d
...
7CCVS



d

Figure 1
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5VCVS



+

Current controlled current source is four
terminal network components that
established a current Icd in the branch of
circuit
...

Unit of transresistance is Volt/Ampere
or Ohm (Ω)
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Current gain is constant
...


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

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Conversely, a current source with a parallel resistor can
be converted into a voltage source with a series resistor
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Source transformation can be applied to dependent source as well
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9Source conversion

Network simplification techniques

+
V1

V2

+

+
-

+ V1 + V2
-

+
-

(a)

+
V1

V2

+

+
-

+ V1 - V2
-

+
-

(V1 > V2)
(b)

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Circuits

Vs

+
-

+ Vs
-

R

R

+

(g)

+

is

is

-

(h)
Figure 1
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4
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
(a)


There are major three electrical circuit elements which are discussed below
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a
l
Figure 1
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12Conductor

Resistance is property of material which opposes the flow current
...

Value of resistance of conductor is
 Proportional to its length
...

 Depends on nature of material
...

l
R
a
ρl
R
a

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The property of the coil of inducing emf due to the changing flux linked with it is known
as inductance of the coil
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 Directly proportional to the area of cross section
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 Depends on absolute permeability of magnetic material
...

Capacitance is the capacity of capacitor to store electric charge
...


Figure 1
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 Inversely proportional to distance between two plates
...


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Current flowing through the conductor is directly proportional to the potential
difference applied to the conductor, provided that no change in temperature
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15Change in current w
...
t change in voltage for conducting material

V I
V  IR


Where R is constant which is called resistance of the conductor
...
g
...

 It cannot be applied to non-metallic conductor e
...
Graphite, Conducting polymers
 It can only be applied in the constant temperature condition
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Circuits
 Branches are meeting at a junction ‘J’
 Incoming current are denoted with (+ve) sign
 Outgoing currents are denoted with (-ve) sign
R2

A

R4

I1 J I3

B

I2
E1

+
-

R3
R5

R1

E2

+
-

D

E

C

Figure 1
...

 Circuit current and loop current are in opposite direction than voltage drop is
denoted by (+ve) sign
...

 If Loop current move through (-ve) to (+ve) terminal of source than direction of
emf is (+ve)
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17Sign convention for Kirchhoff’s voltage law

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18Series combination of resistors
Figure 1
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 Rn

1 1 
1
   
Req  R1 R2 
For n resistor are connected in Parallel

1
1 1 1
1

  
...


PiyushRupala, EE Department



Value of equivalent resistance of parallel
circuit is smaller than the smallest value of
individual resistance of circuit
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D
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Circuits
1
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Explain Voltage divider law and current divider Law
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20Voltage divider circuit

Here , I1  I2  I

V2

Figure 1
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D
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Circuits
1
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Derive the equation of delta to star and star to delta transformation
1

1

R1

R12

R2

R1

R12

R31

R2

R3

R31

R3

2

2

R23

R23
3

3

Figure 1
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22Delta connected network

Resistance between terminal 1 &  2
 R12 (R23  R31 )


Resistance between terminal 1 & 2
 R1  R2

Resistance between terminal 2 & 3

R12 (R23  R31 )
R12  R23  R31

 R2  R3

Resistance between terminal  2 & 3

Resistance between terminal 3 & 1
 R3  R1

 R23 (R12  R31 )


R23 (R12  R31 )
R12  R23  R31

Resistance between terminal 3 & 1 
 R31 (R12  R23 )


R31 (R12  R23 )
R12  R23  R31

Resistance between terminals 1 & 2 in delta equal to resistance between
terminals 1 & 2 in star
R1  R2 

R12 (R23  R31 )
R12  R23  R31

(i )

R2  R3 

R23 (R12  R31 )
R12  R23  R31

(ii )

Similarly ,

R3  R1 

(a)

R31 (R12  R23 )
R12  R23  R31

(iii )

Delta to star conversion

Simplify  i    ii    iii  on both the side of equations

R1  R2  R2  R3 - R3 - R1 

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R12 (R23  R31 ) R23 (R12  R31 ) R31 (R12  R23 )
+
R12  R23  R31 R12  R23  R31 R12  R23  R31

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Circuits
1
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Explain Node analysis
R1

+
V1 -

R3

A

R2

B

R5

+
-

R4

V2
C
Figure 1
...

Node analysis based on Kirchhoff’s current law states that algebraic summation of
currents meeting at junction is zero
...
If there are n nodes in any network,
the number of equation to be solved will be (n-1)
...
Value of
node VC is zero because VC is reference node
...

 Apply the KCL for each node and apply ohm’s law to branch current
...

 Using these voltages, find the required branch currents
...
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

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26Node analysis network for node B

Apply the KCL at node B ,

 I3    I4    I5   0
I3  I4  I5  0
VB - VA VB  VC VB - V2


0
R3
R4
R5
 1
1 1 1 V
VA     VB      2
 R3 
 R3 R4 R5  R5
From, equation (i) & (ii)
1 1 1
R R R
2
3
 1

1


R3




( ii )

 V1 

 
 V
  A    R1 
1 1 1   VB   V2 
  
 
R3 R4 R5 
 R5 


1
R3

One can easily find branch current of this network by solving equation (i) and (ii),if V1 ,
V2 and all resistance value are given
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27Mesh analysis network



Mesh: It is defined as a loop which does not contain any other loops within it
...
D
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Circuits







The current in different meshes are assigned continues path that they do not split at a
junction into a branch currents
...

Steps to be followed in mesh analysis:
 Identify the mesh, assign a direction to it and assign an unknown current in it
...

 Apply the KVL around the mesh and use ohm’s law to express the branch voltage
in term of unknown mesh current and resistance
...

Loop 1
R1

R5

R3
I1

+
V1 -

R2
I1

+
- V2

R4
I2

Figure 1
...
29Mesh analysis network for loop-2

Now Apply the KVL loop  2,
- I2R3 -  I2  I3  R4 -  I2  I1  R2  0
- I2R3 - I2R4 +I3R4  I2R2  I1R2  0

I1R2 - I2  R3  R4  R2  +I3R4  0

R2 I1 -  R3  R4  R2  I2 +R4 I3  0
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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 
  I1 
R2
0

  
R2
R4
 R3  R4  R2 

  I2  

0
R4
  R5  R4    I3 

 -  R1  R2 

R2
0


Δ
R2
R4
 R3  R4  R2 


0
R

R

R
 5 4  
4

 V1

Δ1   0
V
 2

 V1 


 0 
V 
 2 


R2
0

R4
 R3  R4  R2 

R4
  R5  R4  

 -  R1  R2  V1

0


Δ2  
R2
0
R4


0
V

R

R
 5 4  
2

 -  R1  R2 
R2
V1 


Δ3  
R2
 R3  R4  R2  0 

0
R4
V2 

Now ,
Δ
Δ
Δ
I1  1 , I2  2 , I3  3
Δ
Δ
Δ

1
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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
...
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Circuits















According to the application of the superposition theorem
...
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 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)
...

The total current through any portion of the network is equal to the algebraic sum of the
currents produced independently by each source
...

If the individual currents are in the same direction, the resulting current is the sum of
two in the direction of either current
...

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
...

 Replace all the other energy sources by their internal series resistances for
voltage sources
...

 With only one energy source calculate the voltage drops or branch currents
paying attention to the voltage polarities and current directions
...

 Add algebraically the voltage drops or branch currents obtained due to the
individual source to obtain the combined effect of all the sources
...
31Superposition theorem network

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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

R2

A

I3

R3

I4

r

+
V2

B
Figure 1
...
13
...

Thevenin’s can calculate the currents and voltages at any point in a circuit
...
D
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Circuits






Connected across the load“
...

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
...
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
...

Steps to be followed to apply the Thevenin’s theorem:
 Remove the load resistor Rth or component concerned
...

 Find Eth by the usual circuit analysis methods
...



Example network:
R1

r +
V1 -

R2

A

R3

RL

B
Figure 1
...
D
...
Circuits
Step-1
R1

R2

A

+

V1 +
R3

I1

Eth

I2

r
B
Figure 1
...
37Thevenin’s theorem network (step-2)

Step-3
Rth
IL

Eth

+
-

RL

IL 

Eth
Rth  RL

Figure 1
...
D
...
Circuits
1
...
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
...

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
...
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
...

 Find RN by shorting all voltage sources or by open circuiting all the current
sources
...

 Find the current flowing through the load resistor RL
...
40Norton’s theorem network

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Basic Electrical Engineering (3110005)

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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 tan ce
RL  Load Re sis tan ce
Step-2

R1

R2

A
RN 

R3

RN

r

   r+ R  R  + R 
1

3

2

   r+ R1   R3 

RN   
+ R2 

   r+ R1   R3 





B
Figure 1
...
43 Norton’s theorem network (step-3)

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Basic Electrical Engineering (3110005)

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44Charging of capacitor

-

+

C

-

V -

VC

+

C

Figure 1
...
D
...
Circuits
-t

Vc  V (1- e RC )
Also,

dq
dt
d(CVc )
i
dt
-t
d
i  C (V (1- e RC ))
dt
-t
d
i  VC (1- e RC )
dt
-t
  1  RC

i  VC  0-  e


  RC 

i

-t
VC RC
e
RC
V -t
i  e RC
R

i

i  im e

dq
dt
d(CVc )
i
dt
dV
i C c
dt
-t
d
i  C (Ve RC )
dt
-t
-1 RC
i  CV
e
RC
V -t
i  - e RC
R

Also, i 

-t

i  -Ime RC

-t
RC

λ

V
0
...
37

λ

t

Figure 1
...
48Dicharging voltage of capacitor

O

I

t

λ

t

-0
...
37
-Im

λ
Figure 1
...
49Dicharging current of capacitor

1
...
Time domain analysis of first order RL circuit
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51Discharging of inductor

Figure 1
...
D
...
Circuits
λ
0
...
37 Im

t
Figure 1
...
53Dicharging current of inductor

Basic Electrical Engineering (3110005)

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