B.Sc. Sem II AC Circuits

1 | U I , S e m e s t e r - I I ( P A N a g p u r e )
UNIT VI
Alternating current (or voltage)
Definition : The electric current (or voltage) whose magnitude changes with time and its
direction reverses periodically is known as alternating current(or voltage).
An alternating current or a voltage can be express by equations
𝐼 = 𝐼0 𝑠𝑖𝑛𝜔𝑡
Or
𝐸 = 𝐸0 𝑠𝑖𝑛𝜔𝑡
Where I (or E) represent instantaneous current (or voltage)
Io is the peak value of alternating current or amplitude of alternating current;
Eo is the peak value of alternating voltage or amplitude of alternating voltage;
ω is the angular frequency of alternating current, given by
𝜔 =
2𝜋
𝑇
= 2𝜋𝑓; where T is the time period and f is the frequency of alternating current.
Graphically the alternating current or voltage can be represented by the following waveform.
From the above equations it is observe that the alternating voltages or currents are varies with
time 't' as a sin curve, as shown in fig. Therefore they are called as sinusoidal voltage or
sinusoidal current.
The various characteristic parameters associated with the sinusoidal function are given below
-
Amplitude : The maximum or peak value(positive or negative) of an alternating quantity
(voltage or current) is known as its amplitude.
Frequency: The number of cycles per second of a alternating quantity is known as
frequency. It Is denoted by ‘f’. Its unit is Hz.( Hertz ]
Average value of alternating current:
It is the average of all the instantaneous values of the alternating quantity over one
half cycle.
The average value of current can be obtained as follow -
The Instantaneous value of alternating current is given by.
Therefore average value of the current is given by,

𝐼𝑎𝑣 =
1
𝑇/2
∫ 𝐼0 𝑠𝑖𝑛𝜔𝑡
𝑇/2
0
𝑑𝑡
𝐼𝑎𝑣 =
2𝐼0
𝑇
∫ 𝑠𝑖𝑛𝜔𝑡
𝑇/2
0
𝑑𝑡
𝐼𝑎𝑣 =
2𝐼0
𝑇
[−
cos(𝜔𝑡)
𝜔
]
0
𝑇/2
𝐼𝑎𝑣 =
2𝐼0
𝜔𝑇
[cos(𝜔𝑡)] 𝑇/2
0
𝐼𝑎𝑣 =
2𝐼0
𝜔𝑇
[cos(0)− cos( 𝜔𝑇/2)]
As 𝜔 =
2𝜋
𝑇
ℎ𝑒𝑛𝑐𝑒 𝜔𝑇 = 2𝜋 , therefore we get
𝐼𝑎𝑣 =
2𝐼0
2𝜋
[cos(0) − cos( 𝜋)]
𝐼𝑎𝑣 =
𝐼0
𝜋
[1 − (−1)]
𝑰 𝒂𝒗 =
𝟐𝑰 𝟎
𝝅
= 0.636 I0
Thus mean or average value of alternating current during half cycle is 0.636 of its peak value
I0.
Root mean square value (RMS);
“RMS value of alternating current is that value of steady current (dc current) which
produces the same heating effect when flows through a same resister for given time as
alternating current will do in same time.”
OR,
It is the square root of average of all the instantaneous squared values of the alternating
quantity over complete cycle.
If instantaneous value of current is,
Then, mean square value of the current will be,
𝐼2
𝑎𝑣 =
1
𝑇
∫ 𝐼0
2
𝑠𝑖𝑛2
𝜔𝑡
𝑇
0
𝑑𝑡
𝐼2
𝑎𝑣 =
𝐼0
2
𝑇
∫ 𝑠𝑖𝑛2
𝜔𝑡
𝑇
0
𝑑𝑡

𝐼2
𝑎𝑣 =
𝐼0
2
𝑇
∫
1
2
[
𝑇
0
1 − cos(2𝜔𝑡)] 𝑑𝑡
𝐼2
𝑎𝑣 =
𝐼0
2
2𝑇
{∫ 𝑑𝑡 − ∫ cos(2𝜔𝑡)𝑑𝑡
𝑇
0
𝑇
0
}
The cosine function varies between +1 to -1 between the limit of 0 to T , therefore
∫ cos(2𝜔𝑡)𝑑𝑡
𝑇
0
=0
Hence we get
𝐼2
𝑎𝑣 =
𝐼0
2
2𝑇
∫ 𝑑𝑡
𝑇
0
𝐼2
𝑎𝑣 =
𝐼0
2
2𝑇
𝑇 =
𝐼0
2
2
Therefore, 𝐼𝑟𝑚𝑠 = √𝐼2
𝑎𝑣 = √𝐼0
2
2
𝑰 𝒓𝒎𝒔 =
𝑰 𝟎
√ 𝟐
= 𝟎. 𝟕𝟎𝟕 𝑰 𝟎
Thus rms value of alternating current is 0.707 of its peak value I0.
Average Power in ac circuit
If 𝐸 = 𝐸0 𝑠𝑖𝑛𝜔𝑡 be the instantaneous value of emf applied to the circuit and
𝐼 = 𝐼0 𝑠𝑖𝑛( 𝜔𝑡 − ∅) be the instantaneous value of current flowing through the circuit then
instantaneous power consumed in the circuit is given by,
𝑃𝑖𝑛𝑠𝑡 = 𝐸 𝐼 = 𝐸0 𝑠𝑖𝑛𝜔𝑡 𝐼0 𝑠𝑖𝑛( 𝜔𝑡 − ∅)
The average power consumed in the circuit is then given by,
𝑃 =
1
𝑇
∫ 𝐸0 𝑠𝑖𝑛𝜔𝑡 𝐼0 𝑠𝑖𝑛( 𝜔𝑡 − ∅)
𝑇
0
𝑑𝑡
𝑃 =
𝐸0 𝐼0
𝑇
∫ 𝑠𝑖𝑛𝜔𝑡 𝑠𝑖𝑛( 𝜔𝑡 − ∅)
𝑇
0
𝑑𝑡
{since sinA sinB = ½[cos(A-B)-cos(A+B)] }, above equation can be written as,
𝑃 =
𝐸0 𝐼0
𝑇
∫
1
2
[cos(𝜔𝑡 − 𝜔𝑡 + ∅) − 𝑐𝑜𝑠(2𝜔𝑡 − ∅)]
𝑇
0
𝑑𝑡
𝑃 =
𝐸0 𝐼0
𝑇
∫
1
2
[cos(∅)− 𝑐𝑜𝑠(2𝜔𝑡 − ∅)]
𝑇
0
𝑑𝑡

𝑃 =
𝐸0 𝐼0
2𝑇
∫ cos(∅)𝑑𝑡
𝑇
0
− ∫ 𝑐𝑜𝑠(2𝜔𝑡 − ∅) 𝑑𝑡
𝑇
0
Since the cosine function varies between +1 to -1 between the limit of 0 to T, therefore
∫ 𝑐𝑜𝑠(2𝜔𝑡 − ∅) 𝑑𝑡
𝑇
0
= 0
. Hence above equation become,
𝑃 =
𝐸0 𝐼0
2𝑇
∫ cos(∅)𝑑𝑡
𝑇
0
𝑃 =
𝐸0 𝐼0cos(∅)
2𝑇
∫ 𝑑𝑡
𝑇
0
𝑃 =
𝐸0 𝐼0
2𝑇
cos(∅) 𝑇
𝑃 =
𝐸0 𝐼0
2
cos(∅) =
𝐸0
√2
𝐼0
√2
cos(∅)
𝑷 = 𝑬 𝒓𝒎𝒔 𝑰 𝒓𝒎𝒔 𝐜𝐨𝐬(∅), this is the true power
Where𝐸𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 𝑖𝑠 𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑝𝑜𝑤𝑒𝑟 and cos ϕ is called power factor.
J operator:
Let us consider unit vector ( e
r
) along X-axis as shown in figure. Now
rotate this vector through 900 in anticlockwise direction. This rotation
operation takes the unit vector along Y-axis. Let us denote this rotation
operation by ( )J e
r
. Further rotate the unit vector through 900 in
anticlockwise direction. The rotation operation will be now denoted as
2
( ) ( )JJ e J e
r r
which takes it along negative X-axis. i.e, 2
( )J e e 
r r
. Thus
operation of 2
J means multiplication by -1 and hence operation of J means
multiplication by 1 . 1 is imaginary number denoted by j . As single
operation J on any vector ( e
r
) takes it along Y axis ( je
r
), therefore Y-
axis is called as imaginary axis and such plane is called complex plane.
Complex number:
Let us consider a complex quantity given by
Z x jy 
ur
…..(1)
It is represented in complex plane as vector (called as phasor) as
shown in figure. If Z is magnitude of Z
ur
and Z
ur
makes an angle 
with X-axis then
cos ....... (2)
sin ....... (3)
x Z
y Z




From equation (1)
(cos sin )Z Z j  
ur
OR ...... (4)j
Z Ze 

ur
{
*
(cos sin )Z Z j  
ur
OR
* j
Z Ze 

ur
, is called complex conjugate of Z
ur
.}

Squaring and adding equations (2) and (3) we get
2 2
.....(5)Z x y   is called modulus of Z
ur
.
Similarly by dividing equation (3) by (2) we get
tan
y
x
  1
tan .....(6)
y
x
   
   
 
is called phase angle or argument of Z
ur
.
For / 2  , equation (4) becomes
(cos sin )
2 2
Z
j j
Z
 
  
ur
and 2jZ
e
Z


ur
2j
j e 
  similarly, we can show 2j
j e 
  .
AC through pure resistance (R)
Let 0
j t
E E e 
 be the alternating emf connected across pure resistor of resistance R as shown
in figure. If I be the instantaneous value of alternating current flowing through the circuit,
then by Ohms law we get,
𝐸 = 𝐼𝑅
0
j t
E E e IR
 
0 j tE
I e
R

 
0
j t
I I e 
  ……[1]
Where, 0
0
E
I
R

Equation [1] gives instantaneous value of alternating current through pure resistor.
From equation [1] we can say that,
1. Current I is also an alternating with same frequency as that of applied alternating emf
E.
2. There is no phase difference ( = 0) between applied emf E and current I. Both are in
phase in pure resistive circuit.
3. Average power dissipated in the circuit is
cos0rms rms rms rmsP I V I V 
Graphical representation and phasor diagram for emf E and current I is shown in
following figures.
2 2 2
Z x y 
0
j t
E E e 


AC through pure inductor (L)
Let 0
j t
E E e 
 be the alternating emf connected across pure inductor of inductance L as
shown in figure. If I be the instantaneous value of alternating current flowing through the
circuit, then
dI
E L
dt

0
j t dI
E e L
dt


0 j tE
dI e dt
L


Integrating above equation we get,
0 j tE
dI e dt
L

 
0
j t
E e
I
L j



Since 2j
j e 
 , we get
0
2
j t
j
E e
I
L e




2 ( 2)0 0j t j j tE E
I e e e
L L
   
 
 
 
0 0
0Let
L
E E
I
L X
 
( 2)
0
j t
I I e  
 
.................[1]
Here LX L is the inductive reactance [ jωL=jXL is the complex inductive reactance]
Equation [1] gives instantaneous value of alternating current through pure inductor.
E.
2. There is phase difference of
2

between applied emf E and current I. The emf E leads
the current I by
2

in pure inductive circuit. (ELI)
cos 2 0rms rmsP I V  
Hence no power is dissipated in the pure inductive circuit.
0
j t
E E e 


following figures.
AC through pure capacitor (C)
Let 0
j t
E E e 
 be the alternating emf connected across pure capacitor of capacitance C as
shown in figure. If q be the instantaneous charge on the capacitor, then
q
E
C

0
j t q
E e
C


0
j t
q CE e 

Therefore instantaneous value of current is given by
0
j tdq
I CE j e
dt

 
2 ( 2)0 0
1
j j t j t
C
E E
I e e e
C X
   


  
0 0
0Let
1 C
E E
I
C X
 
( 2)
0
j t
I I e  
  ………….[1]
Here 1/ωC is the capacitive reactance [
1
𝑗𝜔𝐶
=
−𝑗
𝜔𝐶
= −𝑗𝑋 𝐶 is the complex capacitive
reactance]
E.
2. There is phase difference of
2

between applied emf E and current I. The current I
leads the emf E by
2

in pure capacitive circuit. (ICE)
cos 2 0rms rmsP I V  
Hence no power is dissipated in the pure capacitive circuit.
0
j t
E E e 


following figures.
AC through series C-R circuit
Let 0
j t
E E e 
 be the alternating emf connected across
resistor R and capacitor C as shown in figure. If I be the
instantaneous value of alternating current flowing through
the circuit, then by applying KVL to the circuit we get,
R C CE V V RI jX I   
Where, C
j
jX
C

  is complex capacitive reactance.
( )CE I R jX 
0
( )
j t
C
E eE
I
R jX Z

 

ur
Where CZ R jX 
ur
is the complex impedance of C-Rcircuit.
As j
Z Ze 

ur
above equation become
0
j t
j
E e
I
Ze



Here 2 2 1 1 1
and tan = tanC
C
X
Z R X
R CR


    
     
  
( )0
2 2
j t
C
E
I e
R X
 
 

( )
0
j t
I I e  
  .................. (1)
Where 0
0 2 2
C
E
I
R X


0
j t
E E e 


Equation [1] gives instantaneous value of alternating current through series C-R circuit.
E.
2. There is phase difference of 1 1 1
tan = tanCX
R CR


    
   
  
between applied emf E
and current I. The current I leads the emf E by in C-R circuit and  depends on
, andC R .
cosrms rmsP I V 
AC through series L-R circuit
Let 0
j t
E E e 
resistor R and inductor L as shown in figure. If I be the
instantaneous value of alternating current flowing through
the circuit, then by applying KVL to the circuit we get,
R L LE V V RI jX I   
Where, LjX j L is complex inductive reactance.
( )LE I R jX 
0
( )
j t
L
E eE
I
R jX Z

 

ur
Where LZ R jX 
ur
is the complex impedance of L-R circuit.
As j
Z Ze 

ur
0
j t
j
E e
I
Ze



Here 2 2 1 1
and tan = tanL
L
X L
Z R X
R R

     
     
  
( )
0
j t
I I e  
  .................. (1)
Where 0
0 2 2
L
E
I
R X


( )0
2 2
j t
L
E
I e
R X
 
 

0
j t
E E e 


Equation [1] gives instantaneous value of alternating current through series L-R circuit.
E.
2. There is phase difference of 1 1
tan = tanLX L
R R

     
   
  
between applied emf E and
current I. The emf E leads the current I by in L-R circuit and  depends on , andL R .
Ex 1: If the voltagein an ac circuit is represented by the equation,v = 220√2 sin(314t – φ) V .
Calculate(a) peakand rms valueof the voltage(b) averagevoltage(c) frequency of AC.
Ex 2: Find the maximum value of current when inductance of two henry is connected to 150V , 50
cycle supply?
Ex 3: A 0.21H inductor and a 12Ω resistance connected in series to a 220V, 50Hz ac source. Calculate
the currentin thecircuit and the phaseanglebetween thecurrent and the sourcevoltage.
Ex 4: A condenser of capacity 10μF connected in series with resistor of resistance 300 Ω and an
alternately voltage of frequency 300/2π Hz is applied across the series combination. Calculate the
phase difference between the voltage and current in the circuit and average power dissipated in the
circuit.
AC through series L-C-R circuit
Let 0
j t
E E e 
 be the alternating emf connected
across resistor R, inductor L and capacitor C as
shown in figure. If I be the instantaneous value of
alternating current flowing through the circuit,
then by applying KVL to the circuit we get,
R L C L CE V V V RI jX I jX I     
Where, LjX j L is complex inductive reactance
and C
j
jX
C

  is complex capacitive reactance.
( )
[ ( )]
L C
L C
E I R jX jX
E I R j X X
  
  
0
[ ( )]
j t
L C
E eE
I
R j X X Z

 
 
ur
Where ( )L CZ R j X X  
ur
is the complex impedance of LCR
circuit.
0
j t
E E e 


As j
Z Ze 

ur
0
j t
j
E e
I
Ze



Here 2 2 1 1
1
( ) and tan = tanL C
L C
L
X X CZ R X X
R R

  
 
  
      
   
 
( )
0
j t
I I e  
  .................. (1)
Where 0
0 2 2
( )L C
E
I
R X X

 
Equation [1] gives instantaneous value of alternating current through series L-R circuit.
E.
2. There is phase difference of 1 1
1
tan = tanL C
L
X X C
R R

  
 
  
   
   
 
between
applied emf E and current I.
Case I:
If L CX X i.e,
1
L
C


 , then net reactance is inductive. In this case the emf E leads the
current I by .
Case II:
If L CX X i.e,
1
L
C


 , then net reactance is capacitive. In this case the current I leads
the emf E by .
Case III:
If L CX X i.e,
1
L
C


 , then the circuit is pure resistive. In this case the current I and
the emf E are in phase.
( )0
2 2
( )
j t
L C
E
I e
R X X
 
 
 

Resonance in series L-C-R circuit
( )L CZ R j X X  
ur
is called the impedance of series LCR
circuit.
Thus, if XL = XC ,Z =R , the impedance of the circuit is minimum and
hence RMS current in the circuit is maximum. This condition is called series
resonance. The frequency at which the resonance takes place is called as
resonant frequency.
At resonant frequency, { &r rf f   }
XL = XC
i.e,
1
r
r
L
C



1
2
2
r
r
f L
f C



2
2
1
4
rf
LC

1
2
rf
LC

This is the expression for resonant frequency.
Voltage across L & C at series resonance:
The voltage across inductor is
𝑉𝐿 = 𝑗𝑋 𝐿 𝐼
Where: 𝐼 =
𝐸
𝑅
is the maximum current through the circuit at resonance.
∴ 𝑉𝐿 = 𝑗𝑋 𝐿
𝐸
𝑅
∴ 𝑉𝐿 = 𝑗
𝑋 𝐿
𝑅
𝐸
∴ 𝑉𝐿 = 𝑗𝑄𝐸
Where: 𝑄 =
𝑋 𝐿
𝑅
=
𝜔𝐿
𝑅
is called quality factor of series resonance circuit.
The voltage across capacitor is
𝑉𝑐 = −𝑗𝑋 𝑐 𝐼
Where: 𝐼 =
𝐸
𝑅
is the maximum current through the circuit at resonance.
∴ 𝑉𝑐 = −𝑗𝑋 𝑐
𝐸
𝑅
∴ 𝑉𝑐 = −𝑗
𝑋 𝑐
𝑅
𝐸
∴ 𝑉𝑐 = −𝑗𝑄𝐸
Where: 𝑄 =
𝑋 𝑐
𝑅
=
1
𝜔𝐶𝑅
is called quality factor of series resonance circuit.
Note − Series resonance LCR circuit is called as voltage magnification circuit,
because the magnitude of voltage across the inductor and the capacitor is equal
to Q times the input sinusoidal voltage E.

Ex. 1: A series resonance network consisting of a resistor of 30Ω, a capacitor of 2uF and an inductor
of 20mH is connected across a sinusoidal supply voltage which has a constant output of 9 volts at all
frequencies. Calculate, the resonant frequency, the current at resonance, the voltage across the
inductorand capacitorat resonance,thequality factorand the bandwidth of thecircuit.
Ex. 2: A series circuit consists of a resistance of 4Ω, an inductance of 500mH and a variable
capacitance connected across a 100V, 50Hz supply. Calculate the capacitance require to produce a
series resonance condition, and the voltages generated across both the inductor and the capacitor at
the pointof resonance.
Ex. 3: Find the value of an inductance which should be connected in series with a capacitor of 5μF, a
resistanceof 10Ω and an ac sourceof 50Hz so thatthe powerfactorof the circuit is unity.
AC through parallel L-C-R circuit
Let 0
j t
E E e 
parallel combination of resistor R, inductor L and capacitor C
as shown in figure. If I be the instantaneous value of
alternating current flowing through the circuit, then by
applying KCL to the circuit we get,
0
j t
E E e 

𝑰 = 𝑰 𝑹 + 𝑰 𝑳 + 𝑰 𝑪
∴ 𝑰 =
𝑬
𝑹
+
𝑬
𝒋𝑿 𝑳
+
𝑬
−𝒋𝑿 𝑪
∴ 𝑰 =
𝑬
𝑹
+
𝑬
𝒋𝑿 𝑳
+
𝑬
−𝒋𝑿 𝑪
∴ 𝑰 = 𝑬(
𝟏
𝑹
−
𝒋
𝑿 𝑳
+
𝒋
𝑿 𝑪
)
∴ 𝑰 = 𝑬(
𝟏
𝑹
− 𝒋[
𝟏
𝑿 𝑪
−
𝟏
𝑿 𝑳
])
Let 𝑌 = (
1
𝑅
− 𝑗[
1
𝑋 𝐶
−
1
𝑋 𝐿
]) =
1
𝑍
be the admittance of parallel LCR circuit then, we have
𝑰 = 𝑬 𝒀
Resonance in parallel L-C-R circuit:
𝑌 = (
1
𝑅
+ 𝑗 [
1
𝑋 𝐶
−
1
𝑋 𝐿
]) =
1
𝑍
is called the admittance of parallel LCR circuit.
Thus, if XL = XC ,Y =1/R , the admittance of the circuit is minimum (impedance is maximum)
and hence RMS current in the circuit is minimum. This condition is called parallel resonance. The
frequency at which the resonance takes place is called as resonant frequency.
At resonant frequency, { &r rf f   }
XL = XC
i.e,
1
r
r
L
C




1
2
2
r
r
f L
f C



2
2
1
4
rf
LC

1
2
rf
LC

This is the expression for resonant frequency.
Current through L & C at parallel resonance:
The voltage across all the elements of parallel RLC circuit at resonance is E = IR.
The current flowing through the inductor
𝑰 𝑳 =
𝑬
𝒋𝑿 𝑳
∴ 𝑰 𝑳 =
𝑰𝑹
𝒋𝑿 𝑳
∴ 𝑰 𝑳 = −𝒋
𝑹
𝑿 𝑳
𝑰
∴ 𝑰 𝑳 = −𝒋𝑸𝑰
Where: 𝑄 =
𝑅
𝑋 𝐿
=
𝑅
𝜔𝐿
is called quality factor of parallel resonance circuit.
The current flowing through the capacitor
𝑰 𝑪 =
𝑬
−𝒋𝑿 𝑪
∴ 𝑰 𝑪 =
𝑰𝑹
−𝒋𝑿 𝑪
∴ 𝑰 𝑪 = 𝒋
𝑹
𝑿 𝑪
𝑰
∴ 𝑰 𝑪 = 𝒋𝑸𝑰
Where: 𝑄 =
𝑅
𝑋 𝐶
= 𝑅𝜔𝐶 is called quality factor of parallel resonance circuit.
Note − Parallel resonance RLC circuit is called as current magnification circuit
because, the magnitude of current flowing through inductor and capacitor is equal
to Q times the input sinusoidal current I.
Ex 1: A parallel resonance network consisting of a resistor of 60Ω, a capacitor of 120uF and an
inductor of 200mH is connected across a sinusoidal supply voltage which has a constant output of
100 volts at all frequencies. Calculate, the resonant frequency, the quality factor and the bandwidth
of the circuit.

Selfinduction:
Production of induced e.m.f. in a coil due to changes of current flowing through the same coil
is called self induction.
If I is the current flowing through the coil and  is the magnetic flux passing through the coil,
then
 α I
Or  =LI ……… (1)
Where L is constant which depends on the nature of the coil. (i.e, number of turns, shape,area,etc)
If the current goes on changing, magnetic flux passing through it also changing and hence
e.m.f. is induced in the coil. This e.m.f. is given by
d
e
dt

  {from eqn (1)}
( )d LI
e
dt
 

dI
e L
dt
 
The negative sign indicates that induce e.m.f. opposes the rate of change of current. The
constant L is called self inductance of a coil.
Mutual induction:
Production of induced e.m.f. in a one coil due to changes of current flowing through the
neighboring coil is called mutual induction.
If I is the current flowing through one coil and  is the magnetic flux passing through the
neighboring coil, then
 α I
Or  =MI ……… (1)
Where M is constant which depends on the nature of two coils. (i.e, number of turns, shape, area,etc)
If the current through one coil goes on changing, magnetic flux passing through the
neighboring coil also changing and hence e.m.f. is induced in the coil. This e.m.f. is given by
d
e
dt

  {from eqn (1)}
( )d MI
e
dt
 

dI
e M
dt
 

The negative sign indicates that induce e.m.f. opposes the rate of change of current. The
constant M is called mutual inductance of a coil.
Transformer:
Transformer is the device used to convert high AC voltage into low AC voltage or vice-versa.
There are two types of transformer,
1. Step down transformer: It convert high AC voltage into low AC voltage.
2. Step up transformer: It convert low AC voltage into high AC voltage.
Construction:
The construction of simple transformer is as show in figure. It consists of
a closed, laminated soft iron core on which two coils having different number of
turns are wound. The coils are made up of insulated wire. The coil to which input
AC voltage is applied is called primary coil and coil across which output AC
voltage is obtained is called secondary.
Working:
For sake of convenience, we make following assumption:
i. Both coils have negligible resistance,
ii. All the magnetic flux linked with each turn of both the coils,
iii. There is no loss of energy in the core.
Such transformer is called ideal transformer.
When AC voltage is applied across the primary, the current through the primary goes on
changing. Due to this, magnetic flux passing through the core goes on changing. As this changing flux
linked with both the coils, an e.m.f. is induced across each coil.
Let N1 and N2 be the number of turns and E1 and E2 be the induced e.m.f. in primary and
secondary coils respectively. If  is flux linked to each turn of both the coils, total flux linked to the
primary coil is 1N  and total flux linked to secondary coil is 2N  . Therefore induced e.m.f. in both
the coils are given by
1 1 1
2 2 2
( )
( )
d d
E N N
dt dt
and
d d
E N N
dt dt




   
   
 2 2
1 1
E N
E N
 …………………..(1)
It can be shown that the e.m.f. induced across primary is equal to the e.m.f. applied between
its terminals.

The ratio 2
1
N
N
is called the turn ratio of the transformer.
From eqn
(1),
If N2 > N1, E2 > E1; so that AC output voltage across the secondary is greater than the AC input
voltage applied across the primary. Such transformer is called Step-Up Transformer.
On the other hand, if N1 > N2, E1 > E2; so that AC output voltage across the secondary is less than the
AC input voltage applied across the primary. Such transformer is called Step-Down Transformer.
Note: for an ideal transformer, efficiency is 100% i.e,
Input power = Output power
I1E1 = I2 E2
Where, I1 and I2 be the current through primary and secondary coil respectively.
 2 1 2
1 2 1
E I N
E I N
  From eqn
(1).
Losses in Transformer
The efficiency of any transformer used in practice is less than 100%. This is due power losses in
practical transformers. The different power losses in a transformer are as given below:
(1) Copper loss:
The primary and secondary coils are usually made up of copper. The coils get heated when
current flows through it, due to Joule’s heating effect. Therefore there is some loss of power as a
heat.
This copper loss can be minimized by using thick wire for coil carrying a high current and thin
wire for coil carrying a low current.
(2) Iron loss: There are power losses in core. These losses are of two types:
(a) Eddy Current loss:
Due to changing currents through the coils, eddy currents are setup in the iron core of the
transformer. Due to this eddy current heat is generated in the coil. Therefore there is some loss of
power.
This power loss can be minimized by using core, made-up of thin laminated sheets, insulated
from each other.
(b) Hysteresis loss:
Due to alternating currents in the coils, iron core suffers magnetization and demagnetization
in opposite directions alternately. The power supplied to the core for magnetization can not be
recovered during demagnetization, because some magnetism retain in the core. This property of
retaining the magnetism is called hysteresis. Due to this hysteresis effect some power is lost in the
core. This loss can be minimized by using the core of materials such as silicon-iron or steel, for
which the hysteresis loss is very small.
(3) Losses due to flux leakage:
All the magnetic flux is not confined to the core. A small fraction passes through air. Thus all
the flux produced by the primary coil does not link to the secondary, hence some power is lost.
This loss can be minimized by designing the transformer in such way that loss is minimum.

Ex 1: A voltage transformer has 1500 turns of wire on its primary coil and 500 turns of wire for its
secondary coil.What will bethe turnsratio (TR) of thetransformer.
Ex 2: If 240 volts rms is applied to the primary winding of the same transformer above, what will be
the resulting secondary no load voltage.

B.Sc. Sem II AC Circuits

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to B.Sc. Sem II AC Circuits

Similar to B.Sc. Sem II AC Circuits (20)

More from Pankaj Nagpure, Shri Shivaji Science College, Amravati

More from Pankaj Nagpure, Shri Shivaji Science College, Amravati (12)

Recently uploaded

Recently uploaded (20)

B.Sc. Sem II AC Circuits