UNIT 2: AC

CIRCUITS
Lecture ALL
Prepared By: Haziqul Yaquin
Sub : ECE249
AC Signal (Sinusoidal Signal or Sine
wave)
 Amplitude
Vm  Phase angle
 Phase Diff
 Time period
 Frequency
AC Signal (Sinusoidal Signal or Sine
wave)
 Amplitude
Vm  Phase
 Phase angle
 Phase Diff
 Time period
 Frequency
 Wavelength
 Half cycle
 Full cycle
Phase and amplitude
Wave forms to phasor diagrams
 Introduction
Electricity supply systems are normally ac (alternating current).
The supply voltage varies sinusoidal
v  Vm sin 2ft 
instantaneous applied voltage,
OR v  Vm sin t 
where
 Vm = peak applied voltage in volts = amplitude
Vm
 f = supply frequency in Hz
 t = time in seconds.

7 7
 Introduction
Electricity supply systems are normally ac (alternating current).
The supply voltage varies sinusoidal
v  Vm sin 2ft 
instantaneous applied voltage,
OR v  Vm sin t 
where
 Vm = peak applied voltage in volts = amplitude
Vm
 f = supply frequency in Hz
 t = time in seconds.

8 8
 Resistance connected to an AC supply

i v
Instantaneous current, i
R
v  Vm sin 2ft  R V
i  m sin 2ft 
R
i  I m sin 2ft 

Current and Voltage are

v
in phase
i

15 15
 Inductance connected to an AC supply
di
vL i – instantaneous current
dt
 Vm ii
v  Vm sin 2ft  i cos2ft    2f
2fL

 Vm  Vm
i cost  Im 
L L
t 0
v  Vm sin 2ft  L
Vm   Current lags Voltage
i sin  t  
L  2 by 90 degree

V V
rms current I 
L 2fL
j
Using complex numbers and the j operator I  V
L
Inductive Reactance X L  2fL  L
V V v

I j 
XL jX L i

16 Phasor diagram and wave form 16

 Capacitance connected to an AC supply
dv i
iC
dt
v  Vm sin 2ft  i  2fCVm cos2ft    2f

v  Vm sin 2ft  C
i  CVm cost 
i
I m  CVm
  Current leads Voltage
i  CVm sin  t  
 2 by 90 degrees

rms current I  CV  2fCV

Using complex numbers and the j operator I   jCV
Capacitance Reactance
1 1
XC   i
2fC C
v
V V V
I j  
XC jX C  jX C 
19 Phasor diagram and wave form 19
 R and L in series with an AC supply i

V  VR  VL
VL

But VR  IR and VL  I  jX L
v  Vm sin 2ft 
VR
 V  IR  jX L 

V V
And I  Where, X L  L  2fL  I
R  jX L R  jL

Complex Impedance Z  R  jL

V R  jL  VR   VL 
Cartesian Form I  I 2  j 2
 R   L   R   L 
2 2 2 2
R  jL R  jL
-j indicates that the current lags the voltage
20 20
  VR   VL 
Complex Impedance: Z  R  jL Cartesian Form: I   2 2 2
 j 2 2 2
R   L  R   L 

In Polar Form phasor diagram constructed with

RMS quantities
V
I ∠-  L  VL

R 2  2 L2

 L  - indicates lagging current.

 L  tan 1   L V
 R  L

I
V
I
R 2  2 L2 VR

 1 L 
Power factor, p.f.  cos  L   cos  tan  Z  R 2   2 L2
 R  jL  jX L
L
Complex impedance: Z  R  jL R

Z  R 2  2 L2
21
21
 RMS and Average Voltage and Current
The “effective” values of voltage and current over the whole cycle

 rms voltage is “RMS value of an alternating current is

V
V m that steady state current (dc) which when
2
flowing through the given resistor for a
given amount of time produces the same
 rms current is Im amount of heat as produced by the
I alternative current when flowing through
2
the same resistance for the same time”

• 0.707 times the maximum value = effective value

Meters normally indicate rms quantities
Average voltage and current
Average voltage is V = 2Vm/pie
Average current is I = 2Im/pie
0.637 times the maximum value
26
 Power Dissipation
We know that: power dissipation | instantaneous = voltage| instantaneous  current | instantaneous

p  vi

Hence, instantaneous voltage, v  Vm sin t 

instantaneous current, i  I m sin t   

Vm I m
p  vi  Vm sin t I m sin t    p cos2t     cos  
2

Vm I m Vm Im
P cos but V & I 
2 2 2

30
Therefore, net power transfer P  VI cos 
30
 Exercise:

For the circuit shown below, calculate the rms current I & phase angle L

0.2H
V
I ∠-  L 
2 2 2
R  L
100V rms
 VR   VL  f = 50 Hz
I 2 j 2 100
 R   L   R   L 
2 2 2 2

 L 
 L  tan 1  
 R 

31 Answer: I = 0.85A -32.10 31

 R and C in series with an AC supply
V  VC  VR But VR  IR and VC  I  jX C 

V i
 V  IR  jX C  I
R  jX C i

VC
1 1 V
but XC    I
C 2fC R   j / C 
v  Vm sin 2ft 
Complex Impedance Z  R  j / C  VR

The current, I in Cartesian form is given by

   
 VR   V / C 
I    j +j signifies that the current leads the
1 1  voltage.
 R2  2 2  R  2 2 
2

32   C    C 
32
    
 VR   V / C 
Complex Impedance: Z  R  j / C  I Cartesian form: I    j
1 1 
 R2  2 2  R  2 2 
2

  C    C 

In Polar Form
V phasor diagram drawn with RMS
I   C +C identifies current leading quantities
2 1 voltage
VR
R  2 2
 C  1 
C  tan 1   I
  CR  C
V
V
I 
2 1
R  2 2
 C VC

Power Factor  cosC  i

VR
Cv
 1  1   v
 cos tan  
  CR   VC
v

33 33
sinusoidal current leading the voltage
 j R
 ZR
C C
1
 jX C   j
C
2 1
Z R  2 2 Z  R2 
1
C  C2
2

34 34
 Exercise:
For the circuit shown, calculate the rms current I & phase angle L
i
V
I   C
1
R2 
 C2
2

0.1F
   
 VR   V / C 
I  j
1  1 
10V rms
R  2 2 
2
 R2  2 2 
  C    C  f = 1000 Hz
1000

Answer: I = 5.32mA 57.90

35 35
 RLC in series with an AC supply
V  VR  VL  VC
We know that: VR  IR VL  I  jX L  VC  I  jX C 

 V  IR  jX L  jX C   IR  jX L  X C 

i
VC

But X L  L & X C  1 / C VC

V VL
I 
R  j  L  1 / C 
v  Vm sin 2ft 
VL

Complex Impedance
 1  1 
2 VVR
Z  R  j   L   
Z  R 2   L  
R

36   C  C  36
 V
From previous page I
R  j  L  1 / C 
VR VL  1 / C  V
I  j I R  jL  1 / C
R 2  L  1 / C  R 2  L  1 / C  R  L  1 / C 
2 2 2 2

V The phasor diagram (and hence the waveforms)

I   s depend on the relative values of L and 1/C.
R  L  1 / C 
2 2
Three cases must be considered

  L  1 / C   X  XC 
s  tan 1   s  tan 1  L 
 R  or  R 

V
I 
R 2  L  1 / C 
2

37 37
 V
From previous page I 
R 2  L  1 / C 
2

(i) L  1 / C VL  VC (ii) L  1 / C VL  VC (iii) L  1 / C VL  VC

VL
VL
VR (VL -VC)

I
VL
I
V V=VR V
VC

I
(VC -VL) VR
VC

VC
capacitive resistive inductive

1
Resonant frequency fo 
2 LC

38 38
 V
From previous page I 
R 2  L  1 / C 
2

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
V
at o, L  1 / C I  0
R
VL = VC and they may be greater than V

1 0 1
0  & f0  
LC 2 2 LC

 fo is called the series resonant frequency.

 This phenomenon of series resonance is utilised in radio tuners.
39 39
 Exercise:
For circuit shown in figure, calculate the current and phase angle and
power factor when frequency is i

(i) 159.2Hz, (ii) 1592.Hz and (iii) 503.3Hz

0.1F

How about you try this ?

100V

1H

1000

Answer:
(i) 11.04 mA + 83.6o, 0.111 leading
(ii) 11.04mA, -83.60, 0.111 lagging
(iii) 100mA, 00, 1.0 (in phase)
40 40
 RMS and Average Voltage and Current
The “effective” values of voltage and current over the whole cycle

 rms voltage is “RMS value of an alternating current is

V
V m that steady state current (dc) which when
2
flowing through the given resistor for a
given amount of time produces the same
 rms current is Im amount of heat as produced by the
I alternative current when flowing through
2
the same resistance for the same time”

• 0.707 times the maximum value = effective value

Meters normally indicate rms quantities
Average voltage and current
Average voltage is V = 2Vm/pie
Average current is I = 2Im/pie
0.637 times the maximum value
41
 Power Dissipation
We know that: power dissipation | instantaneous = voltage| instantaneous  current | instantaneous

p  vi

Hence, instantaneous voltage, v  Vm sin t 

instantaneous current, i  I m sin t   

Vm I m
p  vi  Vm sin t I m sin t    p cos2t     cos  
2

Vm I m Vm Im
P cos but V & I 
2 2 2

42
Therefore, net power transfer P  VI cos 
42
 Exercise:

For the circuit shown below, calculate the rms current I & phase angle L

0.2H
V
I ∠-  L 
2 2 2
R  L
100V rms
 VR   VL  f = 50 Hz
I 2 j 2 100
 R   L   R   L 
2 2 2 2

 L 
 L  tan 1  
 R 

43 Answer: I = 0.85A -32.10 43

 Exercise:
For the circuit shown, calculate the rms current I & phase angle L
i
V
I   C
1
R2 
 C2
2

0.1F
   
 VR   V / C 
I  j
1  1 
10V rms
R  2 2 
2
 R2  2 2 
  C    C  f = 1000 Hz
1000

Answer: I = 5.32mA 57.90

44 44
 Exercise:
For circuit shown in figure, calculate the current and phase angle and
power factor when frequency is i

(i) 159.2Hz, (ii) 1592.Hz and (iii) 503.3Hz

0.1F

How about you try this ?

100V

1H

1000

Answer:
(i) 11.04 mA + 83.6o, 0.111 leading
(ii) 11.04mA, -83.60, 0.111 lagging
(iii) 100mA, 00, 1.0 (in phase)
45 45