Unit content
3. Understand the behaviour of single-phase alternating current (AC) circuits
Series R , L and C alternating current (AC) circuits : current and phase angle in series combinations of RLC circuits (RL, RC, RLC); construction of phasor diagrams and relationship with voltage and impedance triangles for each of the three types of R, L and C combinations; power factor (cos Φ) and power triangle e.g. apparent power (S = VI), true or active power (P =
VI cos Φ) and reactive power (Q = VI sin Φ); conditions for series resonance e.g. inductive reactance equals capacitive reactance (X
L
= X
C
); Q factor (voltage magnification) e.g
Q

V
L
V
, Q

1 L
C
,and its importance in high and low frequency circuits.
R
Parallel : evaluation of the voltage, current and phase angle in parallel combinations of resistance, inductance and capacitance e.g. RL, RC, LC and RLC; construction of phasor diagrams for impedance and phase angle; conditions for parallel resonance in an RLC circuit e.g. supply current and voltage in phase;
Impedance at resonance e.g. dynamic resistance e.g. Q
R
D

L
CR
; Q factor (current magnification)

I
I
C ; filter circuits e.g. high pass, low pass, band pass, band stop.
It assumed that the student has studied the Electrical and Electronic principles module.
© D.J.Dunn www.freestudy.co.uk
1

1. POWER FACTOR CORRECTION
Industrial users of electric power often place inductive loads on the supply in the form of large motors and transformers. This may be regarded as an inductor in series with a resistor and produces a current that lags the supply voltage. The true power is that developed across the resistive load and is given by P = I
2
R and is measured in Watts.
Consider a resistor in series with an inductor. The inductive voltage phasor leads the resistive voltage phasor by 90 o
. The resultant voltage is the supply voltage.
The apparent power is the product of V and I and often stated as
S = VI. This is not measured in Watts but in Volt Amps.
V/I is the impedance Z and is given by Z

R 2 
X 2
The POWER FACTOR is the ratio of the true power to the apparent power and is defined as
P.F.

P

I
2
R

IR

R
 cos
 
hence P = VI cos(

)
S VI V Z
In order to reduce the power factor a series capacitor is needed to produce capacitive reactance equal and opposite to the inductive reactance. Ideally X
C
This is obtained from the relationship X
L
/R = tan

= X
L
Reactive power is defined as Q

X
L
Z

VI sin
WORKED EXAMPLE No. 1
An ac load takes 2.5 kW of power from a supply 110V at 60 Hz. The current is 30 A. Determine the power factor and the size of a capacitor needed in series to correct it.
SOLUTION
P.F. = True Power /Apparent Power = 2500/(110 x 30) = 0.758

= cos
-1
(P.F.) = cos
-1
0.758 = 40.71
o
True Power = I
2
R = 30
2
R = 2500 W
R = 2500/900 = 2.777

X
L
/R = tan

= 0.86
X
L
= 0.86 R = 2.392

X
C
= 2.39 = 1/(2
 fC)
C = 1/(2

x 60 x 2.39) = 0.00111 F
© D.J.Dunn www.freestudy.co.uk
2

SELF ASSESSMENT EXERCISE No. 1
1.
A consumer takes 20 kW of power from an ac supply at 240 V and 50 Hz. Due to an inductive power factor, the current is 100 A. Determine the power factor and the size of a capacitor required to correct it. (0.0024 F)
2.
An electrical load comprises of a resistance of 100

and an inductor of 0.6 H in series. The supply is at 240 V and 50 Hz. Determine the Power factor. (0.47 )
3.
An ac supply to a consumer is at 220V and 50 Hz with a current of 20 A. It is found that there is a lagging phase angle of 20 o
. Determine the Power Factor, the true power and the size of a capacitor that would make the power factor 1.
(0.364, 1.6 kW and 874 μF)
© D.J.Dunn www.freestudy.co.uk
3

2. SERIES CIRCUITS
We know from previous studies that the relationship between current and voltage for any component is related as a ratio X = V/I. For a resistor this ratio is resistance R but for an inductor it is called inductive reactance X
L
and for a capacitor capacitive reactance X
C
.
Inductive reactance increases with frequency and is given by X
L
= 2

f L
Capacitive reactance decreases with frequency and is given by X
C

1
2π f C
When current flows in an RLC circuit, the relationship between it and the resulting voltage is called the IMPEDANCE Z.
Z

V
I
V and I are the resulting r.m.s. volts and current.
Since reactance is V/I it follows that it is also a phasor. The phasor diagram for a series R L C circuit may be drawn as shown with R drawn horizontally to make it easier.
Z
 
X
L

X
C

2 
R
2 and
  tan

1 
X
L

R
X
C
WORKED EXAMPLE No. 2
A resistor of value 470

is connected in series with a capacitor of 22

F and an inductor of 50 mH and a voltage is applied across it. A current of 100 mA (rms) is produced.
Determine the impedance, the phase angle between the voltage and current and the applied voltage when the frequency is 50 Hz
SOLUTION
X
L
= 2πfL = 2π x 50 x 50 x 10
-3
=15.71 Ω
X
C
= 1/2πfC = 1/(2π x 50 x 22 x 10
-6
) =144.6 Ω
Z


X
L

X
C

2 
R
2 

15.71

144 .
6

2 
470
2 
487.4
Ω
  tan

1

X
L

R
X
C
 tan

1
15.71

144
470
V
S
= I Z = 0.1 x 487.4 = 48.7 V rms
.
6
 
15.3
o
© D.J.Dunn www.freestudy.co.uk
4

SELF ASSESSMENT EXERCISE No. 2
1. A resistor of value 4

is connected in series with a capacitor of 47

F and an inductor of 20
μH and a voltage is applied across it. A current of 50 mA (rms) is produced.
Determine the impedance, the phase angle between the voltage and current and the applied voltage when the frequency is 100 Hz. (34 Ω, -83.3
o
and 1.7 V)
2. A resistor of value 0.2

is connected in series with a capacitor of 4.7

F and an inductor of 5 mH and 0.5 V rms is applied across the ends.
Determine the impedance, the phase angle between the voltage and current and the rms current when the frequency is 1000 Hz. (2.455 Ω, -85.3
o
and 204 mA)
3. PARALLEL CIRCUITS.
The main point about parallel circuits is that the voltage is common to each and the current is different. It is often easier to use the idea of ADMITTANCE and CONDUCTANCE so this is defined next.
ADMITTANCE and CONDUCTANCE
These are mainly used in the solution of parallel circuits.
Conductance is the inverse of resistance and is denoted G.
G

1
R
It follows that I
R
= V G
Admittance is the inverse of impedance and is denoted Y.
Y

1
Z
It follows that I = V Y
Susceptance is the inverse of reactance and is denoted B.
B

1
X
It follows that I
C
= V B
C
for a capacitor and I
L
= V B
L
for an inductor.
The units of Y, G and B are Siemens symbol S. 1 S = 1 Ω
-1
PARALLEL RESISTORS
The circuit shows two resistors in parallel. The parallel rule tells us
Z

1
R
1
1

1
R
2
Alternatively
1
Z

1
R
1

1
R
2 or Y

G
1

G
2
© D.J.Dunn www.freestudy.co.uk
5

PARALLEL RESISTOR, INDUCTOR AND CAPACITOR
Consider the parallel circuit below. The voltage is common to all components but the current in the inductor is reactive and leads the voltage by 90 o
and the current in the capacitor lags by 90 o
. Since the current in the resistor is in phase with the voltage, the phasor diagram for the currents is like this.
It is convenient to draw the phasors with I
R
horizontal and the resultant current is as shown.
The resultant current = I


I
L

I
C

2 
I
2
R
The phase angle is
  tan

1 


I
L

R
X
C


If we substitute I
R
Y


B
L

B
C

2
= V G and I
L
= V B
L
and I
C
= V B
C

G
2
  tan

1 
B
L

G
B
C
It also follows that we may represent G, B and Y as phasors.
You should decide which method you prefer to use.
The parallel circuit may be represented with G, B and Y as shown.
WORKED EXAMPLE No. 3
For the three circuits shown below, determine the supply current and phase angle between the supply voltage and current when 50 V rms is applied.
SOLUTIONS
(a) Y


B
L

B
C

2 
G
2 

0

0.2

2 
0.5
2 
0.539
S
I = V
S
x Y = 50 x 0.539 = 26.926 A
  tan

1

B
L

G
B
C
 tan

1
0.2
0.5
 
21.8
o
© D.J.Dunn www.freestudy.co.uk
6

(b) Y


B
L

B
C

2 
G
2 

0.125

0

2 
0.25
2 
0.28
S
I = V
S
x Y = 50 x 0.28 = 13.98 A
  tan

1

B
L

G
B
C
 tan

1
0.125
0.25

0
(c) Y


B
L

B
C

2 
G 2 

26.6
o

0.04

0.05

2 
0.05
2 
0.108
S
I = V
S
x Y = 50 x 0.108 = 5.385 A
  tan

1

B
L

G
B
C
 tan

1
0.04

0.1
0.05

21.8
o
SELF ASSESSMENT EXERCISE No. 3
1.
In a parallel R L C circuit, R = 75 k

, L = 5 μH and C = 0.2 nF. The voltage supply is 0.5 V at
2 Mz. Calculate the admittance, impedance of the circuit, the current and the phase angle.
(0.019 S, 52.9 Ω, 9.4 mA and 45.15
o
)
2.
In a parallel R L C circuit, R = 2 k

, L = 40 mH and C = 20 nF. The voltage supply is 2 V at
100 kHz. Calculate the admittance, impedance of the circuit, the current and the phase angle.
(0.03 S, 33 Ω, 61 mA and 88.9
o
)
© D.J.Dunn www.freestudy.co.uk
7

4. RESONANT CIRCUITS.
SERIES
A series circuit is resonant when the inductive reactance is equal and opposite of the capacitive reactance. It follows that the phase angle is zero. At this condition the reactance is equal to R and is a minimum value. For a given circuit, there will be a frequency f o
where this occurs.
For resonance, X
C
= X
L
2
 f o
L = 1/2
 f o
C where f o
is the resonant frequency.
Rearranging we have:
(2π f o
L)(2π f o
C)

1

2π f
0

2
LC

1 f o
2 
1
(2π )
2
LC f o

2π
1
LC
Q FACTOR
It is quite possible to obtain voltages across a capacitor or inductor larger the supply voltage. We get a magnification. To define this we use the Q factor defined as follows.
Q = V
C
/V for a capacitor and V
L
/V for an inductor. The Q factor may also be based on the current ratio, especially for parallel circuits.
At resonance V = IR since the capacitive and inductive components are equal and opposite so
V
C
= I X
C
= I/2πfC Q
C

1
2 π f o
RC
V
L
= I X
L
= I 2πfL Q
L

2 π f o
L
R
At any other frequency the Q factor is lower and needs to be worked out the hard way. Note that in both cases, the smaller the value of R the larger the Q factor.
Let’s take some typical values V
S
= 10, C = 2mF and L = 2 mH R = 0.1 Ω. The resonant frequency is f o

2π
1
LC

2π
1
2x10
3 x 2 x 10
3

79.6
Hz Q
C
= 1/2
 f o
RC = 10 Q
L
= 2
 f o
L/R = 10
If we calculate V
L
and V
C
over a range of frequencies we get the following result.
We see that the voltages peak at resonance is
100 giving Q = 10 as predicted. If R is zero, then in theory we get an infinite voltage at resonance. If we increase R, we reduce the peak.
© D.J.Dunn www.freestudy.co.uk
8

PARALLE L
Resonance in a parallel circuit occurs when B
L
= B
C
and the resulting admittance is G = 1/R
We will find the result is the same and f o

2π
1
LC
WORKED EXAMPLE No. 4
A series circuit comprises of a resistance of 5 Ω, a capacitor of 2 nF and an inductor of 5 μH.
Calculate the resonant frequency and the current at resonance when 1 V rms is applied.
Calculate the Q factor at resonance.
SOLUTION f o

2π
1
LC

1

1.592
MHz
2π 5 x 10
6 x 2 x 10
9
I = V/Z = V/R = 1/5 = 0.2 A
Q
C
= 1/2
 f o
RC == 1/(2

x 2.59 x 10
6
x 5 x 2 x 10
-9
) = 10
Q
L
= 2
 f o
L/R = 10
SELF ASSESSMENT EXERCISE No. 4
1.
A series circuit has L = 60 mH, R = 15

and C = 15 nF. The supply is 2V ac. Calculate: i.
the resonant frequency (5.3 kHz) ii.
the voltage over each component. (V
R
= 0.133 A, V
C
= 266 V, V
L
= 266 A) iii.
the Q factor for the capacitor and inductor at resonance. (133.3)
2.
A parallel circuit has L = 40 mH, R = 1 k

and C = 10 nF. The supply is 2V ac. Calculate: i.
the resonant frequency (7.96 kHz) ii.
the current in each component.
(I
R
= 2 mA, I
C
= 1 mA, I
L
=1 mA)
© D.J.Dunn www.freestudy.co.uk
9

5. DYNAMIC RESISTANCE
Any inductor is made from a coil of conducting material, usually copper and so has a resistance. We can idealise such an inductor as a resistance in series with pure inductor. This also applies to a series R and L circuit. The phasor diagram is as shown and it follows that the impedance of the coil is:
 
S
2 
R
2 
X
L
2
The admittance of the coil is Y
S

1
Z
S
The phase angle is found from tan(

) = X
L
/R
Now suppose that a capacitor is placed in parallel with the circuit. The phasor diagram using the admittance is shown.
At resonance the inductive and capacitive admittance are equal so the impedance of the circuit is resistive. This is called the DYNAMIC RESISTANCE R
D
.
At resonance capacitive and inductive components are equal so the conductance is G = Y
S
cos(

)
B
C

1
X
C

Y
S sin(

)
Y
S

1
X
C sin(

)
G

1
R
D

Y
S cos(

)
 cos(

)
X
C sin(

)

1
X
C
tan(

)
R
D

X
C
tan(

)

X
C
X
L
R
R
D

2 π f L
2 π f C R

L
CR
WORKED EXAMPLE No. 5
A pure inductance of 150 μH is connected in series with a resistance of 5Ω. A capacitance of
470 nF is connected in parallel as shown. Determine the dynamic resistance and the resonant frequency. If a sinusoidal voltage of 5 V rms is applied at the resonant frequency, what are the rms currents in the inductor and the capacitor? Determine the current Q factor for each.
SOLUTION
R
D

L
CR

150 x10

6
470 x10

9 x 5

63.83
Ω
At resonance X c
= X
L
so f o

2π
1
LC

2π
1
150x10
6 x 470 x 10
9
X
C

1

2 π f o
C
I
C
= 5/17.865 = 0.28 A rms
1
2

(18955)(47 0 x10
9
)

17.865
Ω

18 955 Hz
© D.J.Dunn www.freestudy.co.uk
10

For an inductor and resistor in series
 
2
X
L

2 π f o
L

2

(18955)(15 0 x10
-6
)

R
2 
X
L
2

17.865
Ω This is no surprise because the circuit is resonant.
 
2 
R
2 
Z s
= 18.551 Ω
X
L
2 
5
2 
17 .
865
2
I
L
= 5/18.551 = 0.27 A rms

344 .
149
The impedance of the whole network is the dynamic resistance so I = 5/63.83 = 0.078 A rms
Q
C
= 0.28/0.078 = 3.573
Q
L
= 0.27/0.078 = 3.441
SELF ASSESSMENT EXERCISE No. 5
A pure inductance of 200 μH is connected in series with a resistance of 0.5Ω. A capacitance of
900 nF is connected in parallel as shown. A sinusoidal voltage of 10 V rms is applied at the resonant frequency.
Determine the resonant frequency, the dynamic resistance and the current Q factor for the capacitor and inductor.
(Answers 11 863 Hz, 444.4 Ω, 29.8 and 29.8)
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11

6 FILTERS
You have seen that capacitors and inductors produce reactance that varies with frequency. If we use a suitable circuit containing both we can filter out unwanted components of mixed signals. The diagram shows a LOW and HIGH PASS FILTER .
The high pass filter allows frequencies above a certain value to pass while those below are attenuated. The low pass filter allows low frequencies through but attenuates high frequencies.
Remember capacitive reactance reduces with frequency but inductive reactance increases. The design of the filter depends on the frequencies of interest. One use of a low pass filter is to smooth out direct current with an ac ripple on it and remove mains hum from an audio signal.
Low and high pass filters may be designed with a capacitor and resistor as shown. The time constant RC is important in determining the useful frequencies.
The characteristics of a typical low pass filter are shown. How fast the voltage drops away after the cut off point depends on the design and in particular the Q factor.
Other filters are used to either stop a band of frequencies within a range or either side of a range.
These are called BAND PASS and BAND STOP . The diagram shows a band pass filter design and characteristic. The band width is the frequency range between the cut off point on either side of the filter. The sharpness of the filter depends on the steepness of the sides.
An example of a band pass filter is in radios where the frequencies either side of the frequenc y required are removed. Another example is in signal multiplexing like multiple telephone signals on one line might be removed leaving only the one required.
A band stop filter does the opposite by filtering out a particular range of frequencies. For example a notch filter used in radios will be tuned with a high Q factor to prevent one particular frequency interfering with the rest. The diagram opposite shows a high Q notch filter.
© D.J.Dunn www.freestudy.co.uk
12