Download Analyse series ac circuits

Contents
Introduction
3
Impedance
4
Resistance and inductance in series
6
The practical inductor
6
The voltage triangle
10
The impedance triangle
12
Added series resistance to inductor
16
Resistance and capacitance in series
24
Summary
37
Answers
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EGG202A: 4 Analyse series ac circuits
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Introduction
This section deals with the calculations needed to solve for voltage, current,
impedance and power in series R-L and R-C circuits.
At the end of this section you should be able to:

apply Pythagoras Theorem to a right angle triangle.

define impedance

determine the impedance, current and voltages for a series RL and RC
circuits given the resistance, inductance and supply voltage.

draw and label the an impedance triangle for a series RL and RC
circuits

draw the equivalent circuit of a practical inductor

draw phasor diagrams for a series RL and RC circuits

give examples of inductive and capacitive components in power circuits
and systems and describe their effect on the phase relationship between
voltage and current.

state the AS/NZS 3000 requirements for the installation of capacitors.
EGG202A: 4 Analyse series ac circuits
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3
Impedance
Resistance and reactance are special cases of a more general property called
impedance. You have learnt that:

Ohms Law can be used for resistive ac circuits just as with dc
circuits.

Relations similar to Ohms Law can also be used for capacitors and
inductors. Capacitive reactance XC or inductive reactance XL can be
used instead of resistance in Ohm’s Law.

Pure (ideal) inductors cause the current to lag the voltage by 90°, and
capacitors cause the current to lead the voltage by 90°.
Next we will proceed to analyse circuits that contain both resistive and
reactive components.
Impedance is the property that opposes current in an ac circuit. Impedance is
like a ‘generalised resistance’, and obeys and equation similar to Ohm’s
Law:
Z
V
I
where:
Z = impedance in ohm
V = rms voltage in volt
I = rms current in ampere
Note that:
4

In a resistive circuit, the impedance is equal to the resistance.

In a pure reactive circuit (pure capacitance or inductance), the
impedance is equal to the reactance.

In circuits containing resistance and reactance in series, the
impedance is obtained by combining these two quantities.
EGG202A: 4 Analyse series ac circuits
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For circuits that are series connected we can calculate impedance from the
resistance and reactance. The 90° phase difference between resistor and
reactor voltages allows use of the Pythagorean theorem.
Symbols introduced for the right-angled triangle were:

Z the hypotenuse

X the opposite side to the angle 

R the side adjacent to the angle .
The expression to calculate impedance in a circuit using the right-angled
triangle becomes:
Z  R2  X 2
where:
Z = impedance in ohm
R = resistance in ohm
X = reactance in ohm
The right-angle triangle as shown in Figure 1 is known as the impedance
triangle. Remember that this is used only for series circuits.
Figure 1: Impedance triangle
EGG202A: 4 Analyse series ac circuits
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Resistance and inductance in series
In a pure resistive circuit, the current is in phase with the voltage, and in a
pure inductive circuit, the current lags the voltage by 90°.
Therefore, in an RL circuit, the current will lag the voltage by some angle φ
in the range 0° to 90°.
Figure 2 shows the relationship for the resistive/inductive circuit.
Figure 2: Voltage and current in resistive inductive circuits
The practical inductor
As an inductor is made as a coil of wire it has some resistance even though
this value may be very low. When dealing with inductors in practice both
the resistance and inductance must be considered. We therefore model a real
inductor component using an ideal inductor in series with a resistance. The
circuit model of a practical inductor is shown in Figure 3.
Figure 3: Schematic representation of an inductor
To determine the impedance of an inductor in terms of resistance and
inductive reactance, we can consider the individual voltage drops across
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EGG202A: 4 Analyse series ac circuits
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each equivalent component. These cannot all be measured but can be
calculated
Figure 4: Three voltages to be considered for an inductor
Since this is a series circuit, we take the current as the reference phasor, and
then consider the two voltage drops VR and VXL. These voltages are shown in
the diagram in Figure 5.
The phase relationship of VR and VXL must be considered:

The voltage and current for a resistor are in phase, so VR must be in
phase with the circuit current.

In a pure inductor, the voltage leads the current by 90 degrees, so
VXL leads the circuit current by 90 degrees.
This is shown in Figure 5. Note that VXL therefore leads VR by 90°.
Figure 5: Voltage drops in an inductor
The first step is to measure the dc parameters by applying a DC supply and
measuring the current.
EGG202A: 4 Analyse series ac circuits
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Figure 6: Circuit for measuring DC values
We can now determine the ac parameters by applying a variable ac supply
voltage to give the same current as obtained on dc.
Figure 7: Circuit for measuring ac values
This value is now used in our phasor diagram.
The results for an inductor we have tested are as follows:

10 A dc at 30 V (Figure 6)

ac voltage 50 V (Figure 7) gave 10 A
Using the tip-to-toe phasor diagram, plot the resistive voltage drop
VR = 30 V and the direction of the voltage VXL as in Figure 8.
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Figure 8: Plotting the resistive voltage drop
The applied voltage VAC (50 V) is now included by drawing an arc from the
origin which cuts the line representing the direction of VXL. The VAC phasor
is then drawn between the origin and the point where the arc intersects VXL.
The complete phasor diagram now shows VAC applied, VR and VXL.
Figure 9: Complete phasor diagram for an inductor on ac
The phase angle  can also be measured. The voltage VXL is the
perpendicular phasor and its value, measured from the diagram, is 40 V.
Note that this value could also be determined from Pythogoras’ Theorem.
The final figures for our example are:
VAC = 50 V
VR = 30 V
VXL = 40 V
ϕ = 53.l° with current lagging applied ac voltage
EGG202A: 4 Analyse series ac circuits
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As briefly introduced earlier, the inductor has impedance, resistance and
reactance. From our measurements we can determine these parameters.
Z
Vac
I ac
50
10
 5
V
R  dc
I dc

30
10
 3
V
X L  XL
I
40

10
4

The voltage triangle
In the drawing of the tip-to-toe voltage phasor diagram for an inductor, we
developed a right angle triangle which has the applied voltage on the
hypotenuse, the voltage drop due to resistance on the adjacent side and the
voltage drop due to inductive reactance on the opposite side.
Figure 10: Voltage triangle for inductor
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These three voltage values may be expressed as:
V = IZ
VR= IR
VXL= IXL
The sides of the voltage triangle may be labelled again using these values.
Figure 11: Equivalent voltage triangle
The equivalent voltage triangle is used in the development of the impedance
triangle for a series R-L circuit. The current can be cancelled from each
side, leaving the resistance, reactance and impedance values forming a right
angle triangle.
Student exercise 1
1
Using the voltage triangle find the values VR and VXL if V = 100 V at a 60° (I lagging)
angle.
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Check your answers with those given at the end of the section.
EGG202A: 4 Analyse series ac circuits
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The impedance triangle
If each side of the equivalent voltage triangle of Figure 11 is divided by I,
each side will be labelled as the:

hypotenuse Z

adjacent R

opposite XL.
Figure 12: Impedance triangle for series R-L circuit
Because this triangle no longer represents the voltages, no arrowheads are
used. Each side now represents the electrical properties of:
Z = impedance
R = resistance
XL = inductive reactance
This figure is known as the impedance triangle, and it allows us to calculate
any one of the sides, knowing the other two. To express the impedance of an
inductor in terms of its resistance and reactance, use Pythagoras’s theorem:
From the triangle we can see:
Z2 = R2 + XL2
Impedance Z can then be calculated.
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Z  R2  X L2 ohm
Note: This expression is only for series combinations and should not be
applied to parallel circuit calculations.
The circuit phase angle is maintained in the impedance triangle and is the
angle enclosed by impedance Z and resistance R. This phase angle may be
determined by using either the sine ratio or the cosine ratio.
XL
Z
R
cos  
Z
_ X
  sin 1 L
Z
_ R
  cos 1
Z
sin  
The function sin–1(x) means ‘the angle whose sine is x’ and
the function cos–1(x) means ‘the angle whose cosine is x’. These functions
are known as the ‘inverse sine’ and ‘inverse cosine’ respectively.
The phase angle for any inductor must always be less than 90° due to the
resistance of the inductor winding.
Example
An inductor having a resistance of 47 ohm and an inductance of 1.8 henry is
connected to a 240 V, 50 Hz supply. Determine the:
(a) inductor impedance
(b) circuit current
(c) circuit phase angle.
Draw the phasor diagram.
Solution
(a)
Before we can calculate the impedance, the inductive reactance has
to be calculated.
X L  2 fL
 314 1.8
= 565 
EGG202A: 4 Analyse series ac circuits
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Z  R 2  X L2
 47 2  5652
 567 
(b)
Circuit current
V
Z
240
=
567
 0.423 A
I
(c)
Phase angle
The phase angle has a cosine ratio of resistance over impedance.
R
Z
_ 47
 cos 1
567
_
1
 cos 0.08289
 85.24 lagging (I lags V)
  cos
_
1
Note: it is important that the mode of your calculator is set to degrees when
you calculate sine and cosine, or inverse sin and cosine.
The phasor diagram is shown below:
Figure 13: Using a phasor diagram
Example
An inductor, when tested on a 20 volt dc supply had a 300 mA current, and
when tested on a 20 volt ac supply had a current of 35 mA? Find the
inductance.
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Solution
The calculations to determine the answer use the inductive reactance
expression.
X L  2 fl
L
XL
2 f
To determine XL we can use the impedance expression.
Z  R 2  X L2
X L  Z 2  R2
Finding the values for impedance we have test figures from the ac test.
Z
VAC
I AC
20
0.035
 571 

The resistance is calculated using the DC test figures. We assume here that
the ac resistance is equal to the dc resistance.
R
VAC
I AC
20
0.3
 66.7 

We can now calculate the value of XL.
X L  Z 2  R2
 5712  67 2
 567 
The inductance may now be calculated.
L
XL
2 f
567
2  3.14  50
 1.8 H

EGG202A: 4 Analyse series ac circuits
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Added series resistance to inductor
We will now consider a series circuit which has a resistor and inductor.
A schematic diagram in Figure 14 shows the circuit properties to be
considered.
Figure 14: Resistor in series with inductor
Circuit problems
Given the data in Figure 14, we will be able to solve problems to determine:

inductor impedance and phase angle

circuit impedance, current and phase angle

voltage drops VR andVL to draw the phasor diagram.
Inductor Z:
X L  2 fL
 2  3.14  50  0.5
 157 
Z L  RL 2  X L 2
 302  157 2
 159.8 
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Inductor :
L  cos
_
1
RL
ZL
30
159.8
1
 cos 0.1877
 79.2
 cos 1
Circuit Z:
ZT  RT 2  X L 2
RT  R  RL
 120  30
 150 
ZT  150 2  157 2
= 217 
Circuit I
IT 
V
ZT
240
217
 1.1 A

Circuit 
T  cos
_
1
RT
ZT
150
217
_
1
 cos 0.6912
 46.3 lag
 cos
_
1
Note: The inductor phase angle is 79.2°. With the added resistance the total
circuit phase angle has been reduced to 46.3°.
EGG202A: 4 Analyse series ac circuits
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Phasor diagram
Calculate VR and VL
VR  IR
= 1.1 × 120
= 132 V
VL  IZ L
= 1.1 × 159.8
= 175.78 V
The reference phasor is the current phasor and this need not be drawn to
scale. The scale for the voltages should be 2 V/mm.
Figure 15
Draw reference phasor (see Figure 15). Include VR in phase with I.
132
2
= 66 mm
length of phasor VR =
Include VL (see Figure 16)
176
2
= 88 mm
length of phasor VL =
This voltage leads I by L, that is, 79°
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EGG202A: 4 Analyse series ac circuits
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Figure 16
Figure 17
When the complete phasor diagram has been drawn the total circuit voltage
and phase angle can be measured. Figure 17 shows the length of V to be
120 mm, that is, 240 volt. The phase angle is 46° with current lagging
voltage.
cos  = 0.69
EGG202A: 4 Analyse series ac circuits
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Example
A 100 ohm resistor is connected in series with an inductor. The inductor has
an inductance of 0.95 henry and a resistance of 25 ohm. If the circuit is
connected to a 150 V, 50 Hz supply, calculate:
(a) inductor impedance and phase angle
(b) circuit impedance and phase angle
(c) circuit current
(d) inductor voltage drop and resistor voltage drop.
Draw the circuit phasor diagram tip-to-toe.
Solution
(a)
Inductor impedance and phase angle:
X L  2 fL
 314  0.95
 298.3 
Z L  R 2  X L2
 252  2982
 299 
25
299
_
1
 cos 0.0835
 85.2
L  cos
(b)
_
1
Circuit impedance and phase angle:
Z L  R 2  X L2
where:
RT  R  RL
 100  25
 125 
Therefore:
Z  1252  2982
 323 
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  cos
_
 cos
_
1
1
RT
Z
125
323
_
 cos 1 0.387
 67.24
(c) Circuit current
V
Z
150

323
 0.464 A
I
(d)
Voltage drops:
VR  IR
 0.464 100
 46.4 V
VL  IZ L
 0.464  299
 138.7 V
Figure 18: Circuit diagram for the above example showing V, VR and VL
The circuit phasor diagram (Scale: 1 mm = 2 V) :
EGG202A: 4 Analyse series ac circuits
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Figure 19
The VR phasor is drawn 23 mm in phase with the reference. Tip-to-toe, draw
the voltage phasor for the voltage VL 69 mm at 85° leading the reference.
Draw the applied voltage phasor from the origin to the tip of VL. This, when
measured, is 75 mm and the circuit phase angle can be measured as 67° with
current lagging.
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Student exercise 2
From the data given in Figure 20 answer the following questions.
Figure 20 Rl = 250 ohm, RL = 50 ohm, L = 1 henry
Find the circuit current IT. Draw the phasor diagram to a scale of 2 V/mm.
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2
Calculate the impedance of an inductor if its resistance is 5 ohm and reactance
12 ohm.
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______________________________________________________________________
Check your answers with those given at the end of the section.
EGG202A: 4 Analyse series ac circuits
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Resistance and capacitance in series
Figure 21 below shows a series resistive-capacitive circuit.
Figure 21: Series resistive-capacitive circuit
You should recall that:

In a purely resistive circuit, the current is in phase with the voltage.

In a purely capacitive circuit the current leads the voltage by 90°.
Therefore all ac circuits that have a combination of resistance and
capacitance will have the current leading the voltage by some angle between
0° and 90° as shown in Figure 22. The angle depends on the relative values
of resistance and capacitive reactance in the circuit. The difference between
R-L and R-C circuits is that current leads the voltage in the R-C circuit and
lags in the R-L circuit.
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Figure 22: Phase relationship of an R-C circuit
The selection of the reference phasor depends on whether the voltage or
current is common. In series circuits the current is common to all
components. Current is therefore the reference phasor for series circuits.
A real capacitor component can be considered to be ‘ideal’ because the
resistance is so low it may be neglected.
The phasor treatment for this series RC circuit is very similar to the series
RL circuit:

Firstly, the reference phasor will be the current, as this is common to
all components in a series circuit.

The voltage VR is in phase with the current

The voltage VC lags the current by 90°, so it is plotted downwards on
the diagram.

The supply voltage V is the phasor sum of VR and VC. As these two
are displaced by 90°, total voltage may be calculated using
Pythagoras’ equation.
V  VR 2  VC 2
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The phasor diagram for an R-C series circuit is shown in Figure 23.
Figure 23: Phasor diagram for an R-C series circuit
The angle (is the phase angle between circuit voltage V and current I.
I leads V. This angle may be calculated as the angle whose cosine is the
ratio of VR to V.
  cos
_
1
VR
V
The voltage triangle may now be drawn. This is merely a reproduction of
the phasors of the circuit voltages.
Figure 24: Series R-C circuit voltage triangles
If we now divide each side of the triangle by I, the result is the impedance
triangle. Note that the impedance triangle applies to series circuits only.
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Figure 25: The impedance triangle
Example
A series R-C circuit comprises a 60 ohm resistor and an 80 microfarad
capacitor. The supply is 240 volt, 50 hertz. Determine:
1
circuit impedance
2
circuit current
3
circuit phase angle
4
voltage drops VR and VC
5
draw the tip-to-toe phasor diagram with scale 2 V/mm.
Solution
Figure 26: Phasor diagram
1 Circuit impedance:
First, we require the capacitive reactance:
XC 
1
2 fC
1
2    50  80 106
 39.82 

EGG202A: 4 Analyse series ac circuits
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Therefore the impedance is:
Z  R2  X C 2
 602  39.82
 72 
2 Circuit current:
V
Z
240

72
 3.33 A
I
3 Phase angle:
R
Z
 60 
= cos -1  
 72 
 33.6
  cos
_
1
4 Voltage drops:
VR  IR
 3.33  60
 200 V
VC  IX C
 3.33  39.8
 132.5 V
5 Diagram:
Phasor diagram, with scale 2 V/mm
VR = 200 V  100 mm
V = 240 V  120 mm
VC = 132.5 V  66 mm
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Figure 27: Phasor diagram
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Student exercise 3
Figure 28
For the circuit in Figure 28 determine:
1
circuit Z
____________________________________________________________________
2
circuit I
____________________________________________________________________
3
VR and VC
____________________________________________________________________
4
Draw the phasor diagram with scale 2 V/mm
Check your answers with those given at the end of the section.
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Check your progress
Part A
In questions 1 – 23, place the letter matching your answer in the brackets provided.
1
An ac circuit contains resistance and inductance in series. The phase angle will be
increased if the:
(a) current is increased
(b) voltage is increased
(c) frequency is increased
(d) resistance is increased
2
(
)
The current in a series circuit containing resistance and inductance, with an ac supply,
lags the voltage by:
(a) 90
(b) between 0 and 90
(c) always less than 70
(d) between 90 and 180
3
(
)
(
)
(
)
In a series ac circuit, the ratio of the applied voltage to the current is equal to:
(a) impedance
(b) reluctance
(c) resistance
(d) inductance
4
The phase angle between supply voltage and current, in an ac series circuit,
if resistance and reactance are equal is:
(a) 0
(b) 45
(c) 60
(d) 90
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5
The three sides of an impedance triangle represent:
(a) resistance, reactance and impedance
(b) reactance, inductance and impedance
(c) resistance, inductance and impedance
(d) reactance, inductance and resistance
6
(
)
(
)
(
)
The equivalent circuit of a practical inductor is represented by:
(a) resistor and inductor in series
(b) inductor and capacitor in series
(c) resistor and inductor in parallel
(d) resistor, capacitor and inductor in series
7
An ac circuit contains resistance and capacitance in series. The phase angle will
increase if the:
(a) current is decreased
(b) voltage is increased
(c) frequency is decreased
(d) resistance is increased
8
An ac series circuit contains resistors and a capacitor. You obtain total resistance by:
(a) averaging the resistors
(b) adding the resistors as phasor quantities
(c) finding the sum of the resistor reciprocals
(d) simply adding all the resistor values together
9
(
)
(
)
The current in an ac circuit, consisting of resistance and inductance in series,
lags the applied emf by an angle
(a) of 90
(b) between 0 and 90
(c) between 90 and 180
(d) always less than 60
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10 Current in a series ac circuit with resistance and reactance can be determined from:
(a)
V
Z
(b)
V
R
(c)
V
X
(d)
V
X R
(
)
(
)
(
)
11 Increasing the frequency of the ac supply to a capacitor
(a) reduces the capacitative reactance
(b) increases the capacitative reactance
(c) does not affect the capacitative reactance
(d) varies the capacitance
12 In an impedance triangle the hypotenuse always represents:
(a) impedance
(b) reactance
(c) resistance
(d) reluctance
Part B
1
An inductor when tested is found to have a dc resistance of 5 ohm and an impedance
of 13 ohm at 50 hertz. Determine its inductive reactance and phase angle at 50 hertz.
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2
An inductor having a resistance of 50 ohm and an inductance of 1.5 henry is to be
connected to a 240 volt, 50 hertz supply. What will be the circuit current and phase
angle?
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3
A non-inductive electric furnace is designed to draw a current of 20 ampere from a
240 volt, 50 hertz supply. Determine the inductive reactance of an ideal choke reactor,
connected in series, to provide the same current on a 260 volt, 50 hertz supply.
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4
A series R-L circuit has the following properties:
 the resistor value is 100 ohm
 the inductor has a resistance of 100 ohm and an inductance of 0.25 henry
 supply voltage is 240 volt, 50 hertz.
Calculate the:
(a) total circuit impedance
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(b) total circuit current
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34
EGG202A: 4 Analyse series ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3812
(c) Draw the phasor diagram and determine the phase angle for the inductor alone.
5
Calculate the current drawn from a 240 volt, 50 hertz supply by an 80 microfarad
capacitor.
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6
A series R-C circuit consists of a 150 ohm resistor and a 40 microfarad capacitor
connected to a 240 volt, 50 hertz supply.
Determine:
(a) circuit impedance
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(b) circuit current
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EGG202A: 4 Analyse series ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3812
35
7
An air-cored inductor is connected to a 240 volt dc supply and draws a current of
48 ampere. It is then connected to a 240 volt, 50 hertz supply and draws a current
of 4.8 ampere. What is its reactance at 50 hertz?
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8
An ac circuit has a resistance of 30 ohm connected in series with an inductance of
0.127 henry. When connected to a 230 V supply the current taken is 4.6 A. Determine
the frequency of the supply.
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36
EGG202A: 4 Analyse series ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3812
Summary

The impedance of a series circuit may be calculated using Pythagoras’
equation:
Z  R2  X 2
where:
R  resistance in ohms
X  reactance X C or X L in ohms

Current or in a series R-L or R-C circuit may be calculated using an
equation similar to Ohm’s law.
I
V
Z
where:
V  voltage in rms volts
Z  total impedance in ohms
I  current in rms amperes
EGG202A: 4 Analyse series ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3812
37
Answers
Student exercise 1
1
VR  V cos 
VXL
= 100 × 0.5
= 50 V
 V sin 
= 100 × 0.866
= 86.6 V
Student exercise 2
1
ZL = 318 
L = 81
ZT = 434.3 
IT = 0.553 A
VRI = 138.3 V
VL = 175.8 V
T = 46.3
cos  = 0.6908
2
38
13 ohm
EGG202A: 4 Analyse series ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3812
Figure 29
Student exercise 3
1
50 Ω
2
2A
3
80 V: 60 V
Figure 30
EGG202A: 4 Analyse series ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3812
39
Check your progress
Part A
Part B
1
(c)
1
12 , 67
2
(b)
2
0.506 A, 84
3
(a)
3
5 ,
4
(b)
4
(a) 215 Ω
5
(a)
6
(a)
5
6A
7
(c)
6
(a) 170
8
(d)
9
(a)
10 (a)
(b) 1.12 A
(b) 1.41 A
7
49.7 Ω
8
50 Hz
11 (c)
12 (a)
40
EGG202A: 4 Analyse series ac circuits
 NSW DET 2017 2006/060/04/2017 LRR 3812