Unit2-AC_Circuits.ppt

2
 Most practical applications in electrical engineering involve alternating
current and voltages.
 This unit explains
– analysis of AC circuits and their operations
– use of capacitance transducers.
 After completing this unit you should be able to
– Analyse passive AC circuits comprising resistors, inductors and capacitors
– To determine current flow, voltage distribution and power dissipation.
– Identify series resonance.
– Analyse power factor improvement circuits.
– Describe the operation of capacitance transducers and their use in measuring
displacement.
2

3
21. Introduction
22. Resistance Connected to an AC Supply
23. Inductance Connected to an AC Supply
24. Capacitance Connected to an AC Supply
25. Resistance and Inductance in Series with an AC Supply
26. Resistance and Capacitance in Series with an AC Supply
27. Resistance, Inductance and Capacitance in Series with an AC
Supply
28. AC Supply in Parallel with Capacitance and with Inductance and
Resistance in Series
29. Power Dissipation
30. Capacitance Transducers
31. Problems
3

Vm
4
 Electricity supply systems are normally ac (alternating current).
 The supply voltage varies sinusoidal
 instantaneous applied voltage,  
ft
2
sin
V
v m 

 
t
sin
V
v m 

OR
where
 Vm = peak applied voltage in volts
 f = supply frequency in Hz
 t = time in seconds.
4

5
 
ft
V
v m 
2
sin

i
R
v
i
R
v
i 
Instantaneous current,
 
ft
2
sin
R
V
i m


 
ft
2
sin
I
i m 

Current and Voltage are
in phase
i
Resistance connected to an AC supply
5

6
 The “effective” values of voltage and current over the whole cycle
 rms voltage is
 rms current is
Meters normally indicate rms quantities and this value is
equal to the DC value
Other representations of Voltage or Current are
 maximum or peak value
 average value
2
V
V m

2
I
I m

“RMS value of an alternating current is that steady
state current (dc) which when flowing through the
given resistor for a given amount of time produces the
same amount of heat as produced by the alternative
current when flowing through the same resistance for
the same time”
Root Mean Square (rms) Voltage and Current
6

dt
di
L
v 
 
ft
2
sin
V
v m 
  
ft
2
cos
fL
2
V
i m



 f
2


 
t
cos
L
V
i m




i – instantaneous current
Current lags Voltage
by 90 degree
0
t
m
m
L
V
I 



fL
2
V
L
V
I




rms current
Using complex numbers and the j operator V
L
j
I



L
fL
2
XL 



L
L jX
V
X
V
j
I 


Inductive Reactance





 




2
t
sin
L
V
i m
i
L
 
ft
V
v m 
2
sin

i
v
i
Phasor diagram and wave form
Inductance connected to an AC supply
7
7

i
C
 
ft
V
v m 
2
sin

i
v
dt
dv
C
i 
 
ft
V
v m 
2
sin
  
ft
fCV
i m 
 2
cos
2
 f

 2

 
t
CV
i m 
 cos
 m
m CV
I 

Phasor diagram and wave form
fCV
2
CV
I 



CV
j
I 


Current leads Voltage
by 90 degrees
C
fC
XC


1
2
1


 
C
C
C jX
V
jX
V
X
V
j
I






Capacitance Reactance
rms current








2
sin


 t
CV
i m
i
Using complex numbers and the j operator
Capacitance connected to an AC supply
8
8

9
i
L
V
R
V
 
ft
V
v m 
2
sin

L
R V
V
V 

IR
VR  L
L jX
I
V 

 
L
jX
R
I
V 


L
jX
R
V
I

 fL
2
L
X
,
Where L 



L
j
R
V
I




L
j
R
Z 


L
j
R
L
j
R
L
j
R
V
I







 

















 2
2
2
2
2
2
L
R
L
V
j
L
R
VR
I
Complex Impedance
Cartesian Form
-j indicates that the current lags the voltage
But and
And
R and L in series with an AC supply
9

10
 
L
2
2
2
-
∠
L
R
V
I 








 

 
R
L
tan 1
L
2
2
2
L
R
V
I



L
V
VR
VL
I
L
jX
L
j 

R
2
2
2
L
R
Z 


L
L
j
R
Z 


2
2
2
L
R
Z 


Complex impedance:
  




 


 
R
L
tan
cos
cos 1
L
Power factor, p.f.
-L indicates lagging current.
In Polar Form phasor diagram constructed with
RMS quantities
L
j
R
Z 

 

















 2
2
2
2
2
2
L
R
L
V
j
L
R
VR
I
Complex Impedance: Cartesian Form:
10

11
i
100V rms
f = 50 Hz
0.2H
100
For the circuit shown below, calculate the rms current I & phase angle L





 

 
R
L
tan 1
L


















 2
2
2
2
2
2
L
R
L
V
j
L
R
VR
I
 
L
2
2
2
-
∠
L
R
V
I 



Exercise:
11

12
R
V
C
V
i
 
ft
V
v m 
2
sin

R
C V
V
V 
 IR
VR   
C
C jX
I
V 

 
C
jX
R
I
V 


C
jX
R
V
I


fC
C
XC

 2
1
1


 
C
/
j
R
V
I




 
C
j
R
Z 
/


























2
2
2
2
2
2 1
/
1
C
R
C
V
j
C
R
VR
I



but
Complex Impedance
The current, I in Cartesian form is given by
+j signifies that the current leads the
voltage.
i
But and
R and C in series with an AC supply
12

C
V
VR
VC
I
v
VC
v
VR
v
i
C
C
C
R
V
I 





2
2
2 1






 
CR
C


1
tan 1
2
2
2 1
C
R
V
I



 
C

cos













 
CR

1
tan
cos 1
In Polar Form
+C identifies current
leading voltage
phasor diagram drawn with RMS
quantities
Power Factor
sinusoidal current leading the voltage
 
C
j
R
Z 
/


























2
2
2
2
2
2 1
/
1
C
R
C
V
j
C
R
VR
I



Complex Impedance: I Cartesian form:
13
13

14
C
j
jXC

1



R
2
2
2 1
C
R
Z



C
C
j
R
Z




2
2
2
C
1
R
Z



14

15
0.1F
1000
i
10V rms
f = 1000 Hz
For the circuit shown, calculate the rms current I & phase angle L
Answer: I = 5.32mA 57.90
C
C
R
V
I 





2
2
2 1
























2
2
2
2
2
2 1
/
1
C
R
C
V
j
C
R
VR
I



Exercise:
15

16
R
V
L
V
C
V
i
 
ft
V
v m 
2
sin

C
L
R V
V
V
V 


IR
VR   
L
L jX
I
V   
C
C jX
I
V 

   
 
C
L
C
L X
X
j
R
I
jX
jX
R
I
V 






L
XL 
 C
XC 
/
1

 
C
/
1
L
j
R
V
I

















C
L
j
R
Z


1 2
2 1









C
L
R
Z


Complex Impedance
But &
VC
VL
VR
We know that:
RLC in series with an AC supply
16

17
 
 
 2
2
2
2
C
/
1
L
R
C
/
1
L
V
j
C
/
1
L
R
VR
I













 
 
 
C
/
1
L
j
R
C
/
1
L
R
V
I 2
2









 
s
2
2
C
/
1
L
R
V
I 







 2
2
C
/
1
L
R
V
I





The phasor diagram (and hence the waveforms)
depend on the relative values of L and 1/C.
Three cases must be considered





 



 
R
C
/
1
L
tan 1
s 




 

 
R
X
X
tan C
L
1
s
or
 
C
/
1
L
j
R
V
I





From previous page
17

18
V V=VR V
VR
VR
VC
VL
VC
VL
VC
VL
(VC -VL)
(VL -VC)
I
I
I
(i) C
L 
 /
1
 C
L V
V  (ii) C
L 
 /
1
 C
L V
V  (iii) C
L 
 /
1
 C
L V
V 
capacitive resistive inductive
LC
2
1
fo


Resonant frequency
 2
2
C
/
1
L
R
V
I





From previous page
18

19
From the above equation for the current it is clear that the magnitude of the
current varies with  (and hence frequency, f). This variation is shown in the
graph
C
/
1
L 

 

 0
R
V
I
L
V C
V
at o,
LC
1
0 

LC
2
1
2
f 0
0





 fo is called the series resonant frequency.
 This phenomenon of series resonance is utilised in radio tuners.
and they may be greater than V
=
&
 2
2
C
/
1
L
R
V
I





From previous page
19

20
i
100V
0.1F
1H
1000
For circuit shown in figure, calculate the current and phase angle and
power factor when frequency is
(i) 159.2Hz, (ii) 1592.Hz and (iii) 503.3Hz
(i) 11.04 mA + 83.6o, 0.111 leading
(ii) 11.04mA, -83.60, 0.111 lagging
(iii) 100mA, 00, 1.0 (in phase)
Answer:
How about you try this ?
Exercise:
20

21
VL
VR
VC
ILR
IC
IS
V
LR
C
S I
I
I 
 L
R
C V
V
V
V 


    V
C
/
j
I
jX
I
V C
C
C
C 




 CV
j
IC 

R
I
V LR
R 
   
L
j
I
jX
I
V LR
L
LR
L 


 
L
j
R
I
V LR 

  
L
j
R
L
R
V
L
j
R
V
I 2
2
2
LR 







LR
C
S I
I
I 

IC
IC
IS
ILR
S

V
 
L
j
R
L
R
V
CV
j
I 2
2
2
S 






 
 
L
CL
CR
j
R
L
R
V
I 2
2
2
2
2
2
S 







Can U name the Laws?
We know that:
and
Hence,
Substituting for the different Voltage
components gives:
AC Supply in Parallel with C, and in Series R &L
21

22
240V
IS
100mH
40
C
For the circuit shown calculate the minimum supply current, Is and the
corresponding capacitance C. Frequency is 50 Hz.
Answer: ISmin = 3.71A C = 38.6F
How about you try this one
too?

Exercise:
22

23
power dissipation | instantaneous = voltage| instantaneous  current | instantaneous
i
v
p 


 
t
V
v m 
sin

 

 
 t
I
i m sin
   


 

 t
I
t
V
vi
p m
m sin
sin    
 


 


 cos
2
cos
2
t
I
V
p m
m

cos
2
m
m I
V
P 
2
m
V
V 
2
m
I
I 
 

cos
VI
P 
instantaneous voltage,
instantaneous current,
but &
net power transfer
We know that:
Hence,
Therefore,
Power Dissipation
23


cos
i
i
P
P2
P1
Im
Re
V
P = Apparent power
P1 = Real power
P2 = Reactive power
θ
‫כ‬

sin
i
Real, Apparent and Reactive Power
24
24

i
P
P2
P1
Im
Re
O
V
P = Apparent power
P1 = Real power
P2 = Reactive power
P22
Pn
II
P22= New Reactive Power
Pn= New Apparent Power
I= Current to reduce Reactive Power
I 1
I
Power Factor Correction
25
25

26
 Displacement transducers are often variable capacitors,
 Their capacitance varies with movement.
 The value may be adjusted by varying either
 the distance between the capacitance plates, or
 the effective plate area, or
 the effective dielectric between the plates
fixed plate
moving plate
1. Changing the spacing
d
fixed plate
moving plate
2. Adjusting the plate overlap
deflection
3. Varying the dielectric between the plates
Wedge shaped dielectric
d
A
C r

0

Capacitance
Where
0 = permittivity of free space
r = relative permittivity of dielectric
A = area of overlap between the plates
d = distance between the plates
Capacitance Transducers
26

27
CT
transducer
V
meter
C1
R1
R2
IB
IA
V
To determine the displacement by measuring the
capacitance accurately. When the bridge is
balanced,
1
1
1
C
j
I
R
I B
A



T
B
A
C
j
I
R
I

1
2 

2
1
1
R
R
C
CT 
To achieve the maximum bridge sensitivity:
 the two capacitors should be equal
 the resistances equal to the capacitive reactance at the measuring
frequency.
For accurate measurements prevent or minimise:-
 stray capacitance between leads and earth
 transducer lead inductance
 transducer dielectric losses
 harmonic distortion (undesired components) in voltage supply
27

28
RB
RA
VMeter
displacement
C = C + C C = C - C
V
Linearity of the transducer may be improved by using a
differentially connected displacement device
The transducer is connected to adjacent arms of an ac bridge.
Movement of the central plate increases the capacitance on one side and reduces
it on the other.
28

29
1. AC supply with resistive load, RL in series, RC in
series, RLC in series, and RLC in parallel.
2. Phasor & Cartesian representations.
3. Phase angle and power factor.
4. Dissipated Power.
5. Applications: Capacitance transducer
Conclusion
29

30
Problem Sheet
Q1. A 20V 50Hz supply feeds a 20 Resistor in series with a 100mH inductor.
Calculate the circuit (complex) impedance and current.
Q2. A 200V supply feeds a series circuit comprising 250 resistor, 100mH inductor and
a 159nF capacitor. Calculate the resonant frequency fo and the corresponding
current. Also calculate the current when the frequency is:- fo/3 3fo
Q3. A small company connected to 240V, 50Hz single-phase supply draws a current of
40A at 0.8 power factor lagging. A capacitance is connected across the supply to
improve the power factor of the supply current to:
i) unity ii) 0.95 lagging
Calculate the supply current and capacitance in each case.
Q4. The central plate of a differentially connected displacement transducer shown in
Fig 2.10c is initially midway between the outer plates. Show that if the central
plate is displaced d that the fractional change in the capacitances (C/C) is given
approximately by:
d
d
C
C 


30

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