Worked Examples, Chapter 1
Worked Example 1
Q
A 2 k resistor, a perfect 0.5 H inductor and a perfect 2.2 F capacitor are connected, in turn, across a 5 V,
1 kHz supply. For each case calculate the resulting current flow and sketch the relevant phasor diagram.
A
R 2000 ; L 0.5 H; C 2.2 106 F; V 5 V; f 103 Hz
V
5
volt R
2000
so, I 2.5 mA Ans
Resistor : I V
I
Inductor: Since this is a pure inductor, the only opposition to the flow of current
will be the inductive reactance, XL.
X L 2fL ohm 2 103 0.5
X L 3.142 k V
5
I
amp XL
3142
so, I 1.59 mA Ans
V
I
Capacitor: Similarly, since it is a perfect capacitor, then the only opposition to
current flow is the capacitive reactance, XC.
XC 1
1
ohm 3
2fC
2 10 2.2 106
X C 73.34 k I
V
5
amp XC
73 430
so, I 69.1 A Ans
I
V
151
152
Supplementary Worked Examples
Worked Example 2
Q
A pure inductor is connected across a 10 V, 200 Hz supply, and the current flowing through it is
measured as 0.4 A. Determine the value of its inductance.
A
V 10 V; f 200 Hz; I 0.4 A
V
10
ohm I
0.4
so, X L 25 and, X L 2fL ohm
X
25
so, L L henry 2f
2 200
and, L 19.9 mH Ans
XL Worked Example 3
Q
A perfect capacitor is connected across a 6 V, 5 kHz supply, and the resulting current flow is 88.6 mA.
Calculate the capacitance value.
A
V 6 V; f 5000 Hz; I 88.6 103 A
6
V
ohm 88.6 103
I
so, X C 67.72 1
and, X C ohm
2fC
1
1
so, C farad 2fX C
2 5000 67.72
and C 4.7 107 0.47 F Ans
XC Worked Example 4
Q
A coil of wire is tested by connecting it, in turn, to a d.c. supply and then an a.c. supply. The results
from these two tests are as follows:
d.c. supply of 10 V; resulting current flow 50 mA
a.c. supply of 10 V, 100 Hz; resulting current flow 32 mA
Using the results of these two tests, determine the resistance and inductance values for the coil.
A
d.c. test: V 10 V; I 50 103 A
Supplementary Worked Examples
Since the d.c. current is a steady current then the only opposition to the current
will be the resistance of the coil.
V
10
ohm I
50 103
so, R 200 Ans
R
a.c. test: V 10 V; I 32 103 A; f 100 Hz
In this case the opposition to the flow of alternating current will be the
combined effect of its resistance and its inductive reactance, i.e. the total
opposition is the coil impedance, Z.
V
10
ohm I
32 103
so, Z 312.5 Z
Now, Z R 2 X L2 ohm
or, Z 2 R 2 X L2
so, X L2 Z 2 R 2
and, X L Z 2 R 2 ohm 312.52 2002
57 656
so, X L 240 XL
240
henry 2f
2 100
hence, L 0.382 H Ans
L
Worked Example 5
Q
A coil of resistance 25 and inductance 40 mH is connected to a 50 Hz a.c. supply, and the current
which then flows is 5.36 A. Calculate (a) the supply voltage, (b) the circuit phase angle, and (c) the
power dissipated.
A
R 25 ; L 0.04 H; I 5.36 A; f 50 Hz
(a)
X L 2fL ohm 2 50 0.04
so, X L 12.57 Z R 2 X L2 ohm 252 12.572
and, Z 28 V I Z volt 5.36 28
and, V 150 V Ans
(b) The impedance triangle for the coil is shown below.
XL
Z
φ
R
153
154
Supplementary Worked Examples
φ tan1
XL
R
X
cos1 sin1 L
R
Z
Z
In order to minimise possible errors the last of the above equations will be
avoided, since it involves the use of two previously calculated values. So,
the first equation has been chosen.
XL
12.57
tan1
tan1 0.5028
R
25
and, φ 26.7 or 0.466 rad Ans
φ tan1
P V I cos φ watt 150 5.36 cos 26.7
(c)
so, P 718.3 W Ans
Alternatively, since only resistive components dissipate power, then
P I 2R watt 5.362 25 718.2 W
Note: In this case the power cannot be calculated from P VI watt. This may be
verified by considering the circuit and phasor diagrams as shown below. From
the circuit diagram it can be seen that the p.d. across the resistive component
is VR and NOT V volt. This point illustrates the value of sketching the circuit and
phasor diagrams before proceeding with the calculations.
R
L
25 Ω
40 mH
VL
V
I
VR
VL
V
φ
VR
VR I R volt 5.36 25 134 V
P VR I watt 134 5.36
and, P 718.2 W, wh ich vertifies the previous calculated answer.
Worked Example 6
Q
A 10 µ F capacitor is connected in series with a 270 resistor across a 20 V, 50 Hz supply. Calculate (a)
the current flowing, (b) the p.d.s across the resistor and the capacitor, and (c) the circuit power factor.
A
R 270 ; C 105 F; V 20 V; f 50 Hz
The relevant circuit and phasor diagrams are shown below.
I
Supplementary Worked Examples
R
C
270 Ω
10 µF
VR
VC
φ
VR
I
V
VC
V
20 V
(a)
1
1
ohm 2fC
2 50 105
so, X C 318.3 XC Z R 2 X C2 ohm 2702 318.32
and, Z 417.4 V
20
amp Z
4 17.4
hence, I 47.92 mA Ans
I
(b) VR I R volt 47.92 103 270
VR 12.94 V Ans
VC I X C volt 47.92 103 318.3
VC 15.25 V Ans
(c)
p.f. cos φ R
Z
270
417.4
so, p.f. 0.347 lagging Ans
Worked Example 7
Q
A coil of resistance 330 and inductance 0.25 H is connected in series with a 10 F capacitor. This
circuit is connected across a 100 V, 80 Hz supply. Calculate (a) the circuit current, (b) the p.d.s. across
the coil and the capacitor, (c) the circuit phase angle and power factor, and (d) the power dissipated.
A
R 330 ; L 0.25 H; C 105 F; V 100 V; f 80 Hz
Note that we are dealing with a practical coil, which possesses both resistance
and inductance. In order to simplify the calculations, such a coil is always
considered as comprising a perfect resistor in series with a perfect inductor, as
shown in the circuit diagram below.
I
155
156
Supplementary Worked Examples
coil
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
C
R
L
330 Ω
0.25 H
10 µF
I
VL
VR
VC
Vcoil
V
100 V
(a)
X L 2fL ohm 2 80 0.25
X L 125.66 XC 1
1
ohm 2fC
2 80 105
X C 198.94 Z R 2 (X L X C )2 ohm 3302 (125.66 198.94 )2
and, Z 338 V
100
amp Z
338
hence, I 0.296 A Ans
I
(b)
VR I R volt 0.296 330 97.68 V
VL I X L volt 0.296 125.66 37.2 V
Vcoil VR2 VL2 volt 97.682 37.22
so, Vcoil 104.5 V Ans
Vcoil
VL
VR
I
Alternatively: Z coil R 2 X L2 ohm
3302 125.662
Z coil 353.1 Vcoil I Z coil volt 0.296 353.1
so, Vcoil 104.5 V Ans
VC I X C volt 0.296 198.94
so, VC 58.9 V Ans
XL
Zcoil
R
Supplementary Worked Examples
(c) The complete phasor diagram is shown below.
Vcoil
VL
VR
I
φ
(VC VL)
V
VC
VR
97.68
V
10
hence, p.f. 0.977 lagging Ans
p.f. cos φ phase angle, φ cos1 0.977
so, φ 12.5 lagging Ans
(d) P V I cos φ watt
or P I 2R watt
100 0.296 0.977
P 28.9
9 W Ans
2962 330
P 28.9 W Ans
Worked Example 8
Q
A coil of resistance 500 and inductance 0.2 H is connected in series with a 20 nF capacitor across a
10 V, variable frequency supply. Determine (a) the frequency at which the circuit current will be at its
maximum value, (b) the value of this maximum current, and (c) the p.d.s across both the coil and the
capacitor at this frequency.
A
R 500 ; L 0.2 H; C 20 109 F; V 10 V
For the current to be at its maximum value, the circuit must be supplied at its
resonant frequency, fo Hz. This condition is shown by the phasor diagram below.
VL
Vcoil
VR
VC
I
157
158
Supplementary Worked Examples
(a)
fo 1
2 LC
Hz 1
2 0.2 20 109
hence, fo 2.516 kHz Ans
(b) At resonance, VL VC, so XL XC
so they ‘cancel’ each other
V
10
amp R
500
so, I 20 mA Ans
and I (c)
X C X L 2fo L ohm
2 2516 0.2
so, X C X L 3.162 k VC I X C ohm 0.02 3162
hence, VC 63.23 V Ans
Vcoil I Z coil volt, where Z coil R 2 X L2 ohm
5002 31622
and, Z coil 3201 hence, Vcoil 0.02 3201 64 V Ans