Electrical Engineering

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AC CIRCUITS 18
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COMPARISON OF AC WITH DC
AC DC
1. AC stands for Alternating Current type
supply.
1. DC for Direct Current type supply.
2. Polarity of supply changesperiodically. 2. Polarities are fixed.
3. AC Voltage or current not only varies
its direction but magnitude is varying
with time.
3. Both direction and magnitude are
constant.
4. 4.
5. AC can be transformed from one
voltage level to other voltage level with
the help of transformer.
5. DC cannot be transformed.
6. AC has frequency from low to high,
e.g. mains frequency is 50Hz.
6. Frequency of DC is zero.
7. Examples of AC voltage are Audio
(signal from microphone), AC power
line all oscillator O/P.
7. Dry cell, battery, Rectifier etc.
 ADVANTAGES OF AC
1. The nature of sound is quite similar to that of AC supply and therefore its easy
transmission as well as conversion of sound to electrical signal and electrical
signal to sound with AC; is the main advantages of AC.
2. The transmitted signal easilyreceived by the method of resonance, resonance is
another specialty of AC. This method of receiving radio and TV signal with the
help of resonance is known as „Tuning‟.
3. AC supply can be generated by simple motor without using another electrical
energy. Generator is a common example of AC supply. In vehicles like scooter,
motorcycle, AC magneto is also simple in construction.
AC CIRCUITS

AC CIRCUITS 19
4. DC supply can be generated by using AC. The quality of DC, generated by AC is
not much superior, but that can be filtered by capacitor.
NATURE OF AC
1. Amplitude (A): It is the magnitude of voltage or current at particular instant.
2. Period (T): It is the time taken by AC supply to complete one cycle.
3. Cycle: It is a set of magnitudes in one positive and one negative half cycle.
4. Frequency (f): It specifies the speed of rotating signal in terms of frequency. It is
defined as “frequency is the number of cycle completed in a second”.
1
f Hz
T
 and period
1
T
f
 sec.
5. Wavelength   :
The distance covered by one cycle is called as its wavelength, it is denoted by  (
lambda), and it is measured in meters. The formula for  is

C
f
 Where C = velocity of light = 8
3 10x m/Sec.
6. Peak voltage: It is the maximum value of sine wave voltage either on positive or
negative half cycle.
AC GENERATIOR
As explained earlier AC supply continuously varies in magnitude and periodically
changes polarity. The fig. (a) Shows the idea of AC generated by a rotating loop and
magnitude of voltage for different angular positions. The fig. (b) also shows how a
rotary generator produces and AC voltage. The loop of conductor rotates through the
magnetic field to generate AC voltage across its output terminals. The voltage
generated by this method is known as induced voltage. Its magnitude depends on

AC CIRCUITS 20
the angular position of the loop, at 0
0 it is parallel to magnetic field and at 0
90 , 0
270 it
is perpendicular.When it is parallel,the voltage is zero and at 0
90 it is maximum as
shown in the fig. At 0
180 , 0
270 and 0
360 it passes through same angle but the
direction of current getsreversed because the terminalsof the loop get interchanged.
 RMS value of AC
Ac voltage is measured in terms of effective value which is called as “Root Mean
Square‟ (RMS) Value.RMS value is the most correct value of sine wave voltage.
0.707
2
p
rms P rms
V
Voltage V V x orV 
0.707
2
p
rms p rms
I
Current I I x or I 
 Example 1: Calculate the peak and peak to peak voltage of 230V AC supply.
Solution: 230Vrms V
0.707Vrms Vp x 
0.707
rmsV
Vp 
230
0.707

325.23Vp V
650.46Vp p V 
Thus peak voltage is always more than RMS voltage.
 Example 2: Calculate the RMS voltage of an AC supply whose Peak voltage is
450V.
Solution: 0.707Vrms Vp x 
450 0.707 318.15x V 

AC CIRCUITS 21
REPRESENTATION OF SINE WAVE
If the magnitude of sine wave voltage is noted for 0 0
0 ,90 up to 0
360 , it varies from
0
0 0,
0
90 ,Vp  0 0 0
180 0,270 360 0.Vp and    It is very similar to a sine
function and therefore represented by a sine wave equation
e = A sin ωt and i = A sin t
But w = 2 f ∴ e=Asin2πft
e = A sin t or e = A sin 2 ft
Where „e‟ is AC voltage, A = amplitude of the voltage it is also represented by Emax,
f = the frequency of sine wave
 Phase Angle
(i) If they are passing through the same points as shown in fig. (2.3) then it is said
that these two AC signals are “in phase” or there is 0
0 phase shift.
(ii) If two AC signals are not passing through common points then they are said to be
out of phase signal. The fig. (2.4) shows 0
90 and 0
180 out of phase signals A & B.
It shows that signal voltage A is leading to the signal voltage B. Vector diagram
shows that lengths are equal because their amplitudes are equal.

AC CIRCUITS 22
RESISTANCE IN AC CIRCUIT
Resistance behaves same as it behaves in DC circuit. Ohm‟s law also applicable.
The current flowing through a resistance can be easily calculated by Ohm‟s law
rmsV
i
R
 
 
 
Similarly, it does not affect phase between voltage and current.Therefore resistance
has zero phase angles and it is therefore a non-reactive component.
Note that in AC circuits the resistance value of a resistor remainsunchanged even if
the frequency of AC supply gets changed. In case of capacitor or inductor it makes
difference as explained in the following topic. Hence inductor and capacitor are
known as reactive components.
CAPACITOR IN AC CIRCUITS
Capacitor is a reactive component it behaves in different way in AC circuit. When a
capacitor is connected in AC circuit, it‟s resistance to AC, which is called as
“reactance” it depends upon the frequency of AC voltage. Resistance in DC circuit
and reactance in AC circuit, these two terms are different. Reactance is also
measured in Ohms but it is not constant, it varies with frequency. “Xc” denotes
capacitive reactance and it depends upon the value of capacitor as well as frequency
of AC supply. Hence it is calculated by formula.
1
2
Xc
fc
 
Similarly, capacitor produces 0
90 phase shift between voltage and current as shown
in fig (2.8), current leads voltage by 0
90

AC CIRCUITS 23
Refer fig. (2.8) when capacitor is connected across a DC supply the lamp glows for a
very short time and then it becomes OFF because current through the circuit
becomes zero once capacitor is fully charged it is said that capacitor is DC open.
In the next figure; when it is connected across an AC supply lamp glow due to
charging and discharging, it is said that capacitor is AC short. When the frequency
of AC supply is increased the lamp glows with more intensity. It shows that the
capacitive reactance decreases with increase in frequency.
INDUCTOR IN AC CIRCUIT
Inductor is nothing but a coil it is also a reactive component, its opposition, to AC
current is called as “inductive reactance” denoted by  LX and it is calculated by
formula,
2LX fL 
Measured in Ohms. Inductive reactance is directly proportional to the value of
inductance and frequency.Inductor also produces
0
90 phase shift between voltage
and current but here current lags voltage by 0
90 as shown in fig (2.9).
Refer fig. ( 2.9) when an inductor is connected across DC supply the lamp glows
continuously because the inductor is acting as a wire, it is said that inductor is DC
short, When it is connected across an AC supply lamp does not glow because it
opposes AC current, it is said that inductor is AC open.

AC CIRCUITS 24
IMPEDANCE (Z)
When circuit is complex, if capacitor, inductor and resistor all are present then
total opposition is not the sum of Xc, LX and r but it is calculated by other method
because there is an effect on phase shift. This effective total opposition is called as
“impedance”. It is denoted by “Z”. When circuit is series circuit then impedance is
calculated by
2 2
Z X R 
Consider its vector diagram. If the phase of R is taken as a reference then the phase
of reactance is 0
90 out of phase. In case of LX it makes 0
90 phase shift while in
case of CX it makes phase shift of 0
90 .
2 2
Z X R 
While 1 1
tan tan tanC CL LX XX X
While or or
R R R R
    
 
X is the total reactance and R is the total resistance of the circuit. If circuit is iin
parallel then impedance is calculated by current equation
2 2
T X RI I I 
Where x L CI I I
and therefor T
V
Z
I

Example 2 Calculate impedance of the circuit

AC CIRCUITS 25
Solution:
As shown in given circuit diagram components are in parallel therefore first total
current  TI must be calculated.
100 100 100
4 1 4
25 100 25
L C R
L C
V V V
I A I A I A
X X R
         
(Voltage across each is 100V because they are in parallel)
4 1 3XI A A A    (Capacitive and Inductive currents are 0
180 out of phase)
2 2 2 2
3 4 5
100
20
5
T X R
T
I I I A
V
Now impedance Z
I
    
   
COMPARISON OF CX AND LX
Capacitive Reactance ( CX ) Inductive Reactance ( )LX
1. It is the opposition of capacitance
to AC current.
1. It is the opposition of inductance to
AC current.
2. Formula : 1/ 2CX f C 2. 2LX f L
3. It increases for lower values of
capacitors.
3. It decreases for lower values of
inductors.
4. It increases for lower frequencies. 4. It increases for higher frequencies.
5. Voltage lags to the current by 0
90 5. Voltage leads to the current by 0
90
6. 6.
7. 0
lags by90C CV i 7. 0
leads by90L LV i

AC CIRCUITS 26
SOLVED PROBLEMS
1. If the value of peak voltage is 325.25 V find its RMS value
Solution
 
 
 
 
0.707 log325.25 log0.707
325.25 0.707 2.5122173 ( 0.155805)
229.95 2.3616
log(2.3616) 229.5
Vrms Vp
Vrms volt
Anti
   
    
 
 
2. Find the impedance when resistance of 30  and inductance of 40  are in
series.
Solution
   
2 22 2
330 40
ImpedanceZ 40 30 50
R X
X R
   
     
3. Resistance of 20 ohms, inductor having inductive reactance of 20 and capacitor
having capacitive reactance of 60. If they are connected in series find impedance
and phase angle of the circuit.
Solution
   
2 2
2 2
1 0
Given R = 20 ,X 20 , 60
60 20 40
40 20 44.72
20
tan 1
20
tan 1 45
L C
C L
L
X
Seriescircuit formulais Z X R
Where X X X
Z
X
R

 
   
 
     
    
  
  
RESONANCE (TUNED CIRCUIT)
This is a useful term; it plays in an important role in radio, TV circuit to tune or to
select p articular station signals. Resonance is defined, as “it is the moment in
electrical circuit where circuit gives maximum response at particular frequency”.
Response may be impedance or current in the circuit. The frequency at which
circuit becomesresonant is called as „resonant‟ frequency and it is denoted by ( )rf
similarly the circuit, which shows resonance is known as tuned circuit. The main
use of this circuit we find in radio and TV circuits. In electrical L-C-R circuit
inductive reactance increases with frequency and capacitive reactance decreases. At

AC CIRCUITS 27
a particular frequency LX becomes equal to CX they are equal and opposite this
phenomenon is known as resonance. The frequency at which it happens is known as
resonant frequency (fr.)
There are two resonance circuits as follows
1. Series Resonance and
2. Parallel Resonance.
1. Series Resonance
A capacitor, inductor and AC voltage source when they are connected in series
then they form series resonance circuit.
shows the circuit diagramwhere „R‟ is an internal (DC) resistance of the inductor
therefore it is also known as LCR circuit. The current meter shows the current for
different frequenciesof AC voltage source. The graph shows at resonant frequency
„ rf ‟current is maximum and above „ rf ‟ or below „ rf ‟ current is rapidly dropping.
Derivation
2
2
At Resonance
1
2
2
1
4
1
......................Formula for Resonant frequency.
2
L C
r
r
r
r
X X
fr L
f C
f
LC
f
LC





 
 
 

AC CIRCUITS 28
 Characteristics Of Series Resonance
1. Below resonant frequency; circuit is capacitive and current is small due to high
CX
2. Above resonant frequency; circuit is inductive and current is small due to high
LX .
3. At resonance circuit is resistive and maximum current flows through circuit
because 0.L CX X 
4. At resonance voltage across capacitor is equal to the voltage across inductance
but they are in 0
180 out of phase. C L C LI x X I x X E E   because series
current I is same and at resonance C LX X
5. At resonance, impedance of the circuit is very small.
6. The circuit is known as an acceptor circuit.
Example
A series resonance circuit consists of 100 micro Henry inductance and 100pf
capacitance. Find the resonant frequency.
Solution
C = 100 pf L = 100 H
 
6 4 18
714/214
7 3
1
:
2
1 1
2 3.14 100 10 100 10 12 6.28 10 10
1 1 1
6.28 106.28 106.28 10
0.1592 10 1592 10 1592
rFormula f
LC
X X X X X X
X
X X Hz or KHz

  


 
 
 
 ‘Q’ of the Resonance
Illustratesthe idea of q and its relationship with „Bandwidth‟. The bandwidth of
resonance is important because it gives the idea about the band of frequencies for
which resonance is effective. As shown in fig (2.13) bandwidth is calculated by
formula,
10 .
( ) 0.2 200
50
MHz
Bandwidth f MHzor KHz  
It shows 0.1 MHz below and 0.1 MHz above from 9.9 MHz-10 MHz-10.12 MHz
actual resonances occur. Current for these frequencies is approximately close to
the maximum value. Therefore this resonance circuit is called as „Band-pass-
filter‟ it gives maximum output for a band of frequencies, For example all other
frequencies above 10.1 MHz and below 9.9 MHz output is about zero, those
frequencies are filtered. This series resonance is also known as an acceptor
circuit.

AC CIRCUITS 29
2. Parallel Resonance
This is another L-C-R circuit in which L-C are connected in parallel with input
AC voltage. The basic difference between series and parallel resonance is, in
parallel resonance impedance of circuit is maximum while in series resonance
current is maximum. Resonant frequency is calculated by formula
1
2
rf
LC

Because at this frequency capacitive reactance becomes equal to the inductive
reactance ( ).C LX X
 Characteristics Of Parallel Resonance
1. Below resonant frequency; the circuit is inductive and impedance is small
because LX is low.
2. Above resonant frequency; the circuit is capacitive and impedance is small
because CX is low.
3. At resonance; the circuit is resistive and impedance is maximum because
L CX X
4. The circuit is known as rejecter circuit.
COMPARISON
Series Resonance Parallel Resonance
1. L and C are in series 1.L and C are in parallel.
2. At resonance, current is maximum 2.At resonance, current is minimum
3. At resonance impedance in
minimum
3.At resonance impedance is maximum
4. Circuit is capacitive below rf and
inductive above „ rf ‟
4.Inductive below rf and capacitive above
rf .
5. Known as Band pass filter or
acceptor circuit
5.Known as band-stop filter or rejecter
circuit.
6. Formula :
1
2
rf
LC
 6.Formula :
1
2
rf
LC


AC CIRCUITS 30
SOLVED PROBLEMS
1. The Resonant frequency of a LC circuit is 1 KHz calculate the value of
inductance, if the capacitance is 50 f
Solution Given fr = 1 KHz C = 50 f .
3
6
6
2 6
4
3
1
2
1
1 10
6.28 50 10
squaring bothsides
1 1 1
1 10
(6.28) 50 10 39.43 50 1971.92
5.07 10
0.5 10
0.5
fr
LC
X
lx x
X
L
H
mH





 

  
   
 
 

2. A capacitor of 50 ,f an inductance of 0.2025 H and a resistance of 21 are
connected in series.At what frequency will the resonance occur? What will be the
current at resonance if the supply voltage is 14 volts?
Solution (I)
2
2 2 6
3 6 3
6
2
1
2
1 1
4 4(3.14) 0.2025 50 10
1
2.5042 10 10 2.5042 10
399.31 10
2504.2
50.04
r
r
fr
LC
fr
LC
f Hz
f Hz

 



  
  
     


 
(II) At resonance circuit resistance is only the resistance of coil, which is given
as 21 
Current in this circuit is
14
0.66
21
V
i A
R
   

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