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HNC Electrical and Electronic Engineering
Year Two - 2014/15
Module: Electronics

ELECTRICAL & ELECTRONIC

PRINCIPLES

AC THEORY

Keith A
...
Hudson

2

Electrical & Electronic Principles
06 November, 2014

Contents
1

Question 1
...
5
Voltage across 𝒁𝟏 (Q1b)
...
7
Power used by each impedance (Q1d)
...
8
Thévenin Equivalent Circuit (Q2a)
...
10
Current through load (Q2c)
...
13
Delta (Δ) configuration
...
14
Wye (Y) configuration
...
17
Inductance and capacitance (Q4a,b)
...
18
Impedance at half-power frequencies (Q4e)
...
20
Resonant Frequency (Q5a)
...
20
Dynamic Resistance, Supply Current and Supply Voltage
...
21
Capacitor Branch
...
22
Current in the Inductor Branch (Q5c)
...
24

7

Question 7
...
25
Coils in Series (Q7b)
...
29

HNC Electrical and Electronic Engineering

Year Two: 2014/15

Keith A
...
5
Figure 2: Circuit for Q2
...
8
Figure 4: Thévenin equivalent circuit
...
11
Figure 6: Load replaced
...
13
Figure 8: circuit with nodes and impedances named
...
14
Figure 10: Circuit with Δ replaced by Y equivalent
...
20
Figure 12: Resonant frequency equation
...
21
Figure 14: Circuit for Q6
...
25
Figure 16: Dot notation with respect to windings (Potisuk, 2014)
...
26

HNC Electrical and Electronic Engineering

Year Two: 2014/15

Keith A
...
5
Table 2: Calculation of current in each branch
...
6
Table 4: Calculation of Total Current
...
7
Table 6: Thévenin equivalent calculations
...
10
Table 8: Current through load
...
15
Table 10: Known values
...
17
Table 12: Bandwidth and half-power frequencies
...
18
Table 14: Upper half-power frequency impedance calculations
...
20
Table 16: Dynamic resistance and inductor current calculations
...
21
Table 18: Capacitor current calculation
...
22
Table 20: Current through the inductor
...
27
Table 22:
...
Hudson

Electrical & Electronic Principles
06 November, 2014

5

1 Question 1

Figure 1: Circuit for Q1
Table 1: Known values

Polar Form

Rectangular Form

𝑆𝑢𝑝𝑝𝑙𝑦 𝑉𝑜𝑙𝑡𝑎𝑔𝑒, 𝑉 𝑆 = (200 + 𝑗0) 𝑉

𝑉 𝑆 = 200 ∠0° 𝑉

𝑍1 = (10 − 𝑗45) 𝛺

𝑍1 = 46
...
47119229° 𝛺

𝑍2 = (12 − 𝑗30) 𝛺

𝑍2 = 32
...
19859051° 𝛺

𝑍3 = (30 + 𝑗40) 𝛺

𝑍3 = 50 ∠53
...
Given the supply voltage, we can then
calculate the current in each branch
...
Hudson

Electrical & Electronic Principles
06 November, 2014

6

Table 2: Calculation of current in each branch

𝑍12 = 𝑍1 + 𝑍2
𝑍12 = (10 − 𝑗45) + (12 − 𝑗30)
𝑍12 = (10 + 12) + (−𝑗45 − 𝑗30)
𝑍12 = (22 − 𝑗75) 𝛺

𝑍12 = 78
...
65182845° 𝛺

𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑍1 , 𝑍2 𝑖𝑠 𝐼12 =

𝑉𝑆
𝑍12

200 ∠0°

𝐼12 = 78
...
65182845°
200

𝐼12 = (78
...
65182845°)
𝐼12 = 2
...
65182845° 𝐴

𝐼12 = (0
...
455393682) 𝐴

𝐼12 = 2
...
65° 𝐴 (2𝑑𝑝)

𝐼12 = (0
...
46) 𝐴 (2𝑑𝑝)

𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑍3 , 𝐼3 =

𝑉𝑆
𝑍3

200 ∠0°

𝐼3 = 50 ∠53
...
13010235°)
𝐼3 = 40 ∠(−53
...
13°) 𝐴 (2𝑑𝑝)

𝐼3 = (24 − 𝑗32) 𝐴

Voltage across 𝒁 𝟏

(Q1b)

Table 3: Calculation of voltage across 𝒁 𝟏

𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑍1 , 𝑉1 = 𝐼12 𝑍1
𝑉1 = 2
...
65182845° × 46
...
47119229°
𝑉1 = (2
...
09772229) ∠(73
...
47119229°)

𝑉1 = 117
...
81936384° 𝑉

𝑉1 = (117
...
857259788) 𝑉

𝑉1 = 117
...
82° 𝑉 (2𝑑𝑝)

𝑉1 = (117
...
86) 𝑉 (2𝑑𝑝)

HNC Electrical and Electronic Engineering

Year Two: 2014/15

Keith A
...
720248813 + 𝑗2
...
720248813 + 24) + ( 𝑗2
...
72024881 − 𝑗29
...
52238912∠ − 50
...
72 − 𝑗29
...
52 ∠ − 50
...
)
𝐼12 = 0
...
720248813)2 × 10
𝑃1 = 5
...
19 W (2dp)

𝑃2 = 𝑃𝑜𝑤𝑒𝑟 𝑖𝑛 𝑍2 , 𝑃 = 𝐼 2 𝑍 (We are only interested in the Real parts of the rectangular notation
...
720248813, 𝑍2 = 10
𝑃2 = (𝐼12 )2 × 𝑍2
𝑃2 = (0
...
225100232 W
𝑃2 = 6
...
)
𝐼3 = 24, 𝑍3 = 30
𝑃1 = (𝐼3 )2 × 𝑍3
𝑃1 = (24)2 × 30
𝑃1 = 17280 W (I have no idea why this figure is so much higher than the others
...
Hudson

8

Electrical & Electronic Principles
06 November, 2014

2 Question 2

Figure 2: Circuit for Q2

Thévenin Equivalent Circuit

(Q2a)

Step 1: Remove load - Already done
...
See Table 6
...
See Table 6
...
Hudson

Electrical & Electronic Principles
06 November, 2014

9

Table 6: Thévenin equivalent calculations
Polar: (r ∠θ)

Polar-->Rectangular

Rectangular: (x+jy)

Rectangular-->Polar

θ

x=r cosθ

y=r sinθ

x

𝑟 = √𝑥 2 + 𝑦 2

𝑦
𝜃 = tan−1 ( )
𝑥

"=SQRT(G5^2 + H5^2)"

r

"=DEGREES(ATAN2(G5,H5))"

y

Vs

9
...
000000000000

9
...
000000000000

Z1

0
...
000000000000

10
...
000000000000

Z2

12
...
000000000000

12
...
000000000000

Z1 + Z2

12
...
000000000000

15
...
805571092265

x

y

r

θ

VTH=Vs*Z2/(Z1+Z2)

6
...
805571092265

5
...
426229508197

ZTH=Z1*Z2/(Z1+Z2)

7
...
194428907735

4
...
901639344262

r

θ

x

y

HNC Electrical and Electronic Engineering

Year Two: 2014/15

Keith A
...
913991516376

-39
...
311475409836

-4
...
682212795974

50
...
918032786885

5
...
682212795974

50
...
918032786885

5
...
900000000000

-90
...
000000000000

-0
...
Hudson

11

Electrical & Electronic Principles
06 November, 2014

Figure 5: Norton equivalent circuit

Current through load

(Q2c)

Figure 6: Load replaced

HNC Electrical and Electronic Engineering

Year Two: 2014/15

Keith A
...
155494421404

-21
...
000000000000

-6
...
918032786885

-15
...
156016827130

-5
...
347122207818

-23
...
317343611272

-0
...
Hudson

Electrical & Electronic Principles
06 November, 2014

13

3 Question 3

Figure 7: Circuit for Q3

To calculate the total circuit impedance for Figure 7 we must first convert it to a form where serial and parallel
impedances can be calculated
...
5)Ω
d
Figure 8: circuit with nodes and impedances named

HNC Electrical and Electronic Engineering

Year Two: 2014/15

Keith A
...


Figure 9: Circuit partially re-arranged in a delta format

Convert from delta (Δ) to Wye (Y)
...
All calculations are shown in Table 9
...
Hudson

Electrical & Electronic Principles
06 November, 2014

15

Table 9: Impedance calculations
Polar: (r ∠θ)

Polar-->Rectangular

Rectangular: (x+jy)

Rectangular-->Polar

θ

x=r cosθ

y=r sinθ

x

y

𝑟 = √𝑥 2 + 𝑦 2

𝑦
𝜃 = tan−1 ( )
𝑥

"=SQRT(G5^2 + H5^2)"

r

"=DEGREES(ATAN2(G5,H5))"

Zab

0
...
000000000000

5
...
000000000000

Zac

0
...
000000000000

10
...
000000000000

Zbc

0
...
000000000000

5
...
000000000000

Zab + Zac + Zbc

0
...
000000000000

20
...
000000000000

Zbd

0
...
500000000000

22
...
000000000000

Zcd

5
...
000000000000

5
...
000000000000

Z1 = Branch e-c-d = Zc + Zcd

5
...
500000000000

5
...
565051177078

Z2 = Branch e-b-d = Zb + Zbd

0
...
250000000000

21
...
000000000000

0
...
032941176471

0
...
633633998940

5
...
734439834025

7
...
916193087857

6
...
73

7
...
92

Zab * Zac

50
...
000000000000

-50
...
000000000000

Za = (Zab * Zac) / (Zab + Zac + Zbc)

2
...
000000000000

0
...
500000000000

Zab * Zbc

25
...
000000000000

-25
...
000000000000

Zb = (Zab * Zbc) / (Zab + Zac + Zbc)

1
...
000000000000

0
...
250000000000

Zac * Zbc

50
...
000000000000

-50
...
000000000000

Zc = (Zac * Zbc) / (Zab + Zac + Zbc)

2
...
000000000000

0
...
500000000000

1 / Z1

0
...
565051177078

0
...
080000000000

1 / Z2

0
...
000000000000

0
...
047058823529

1 / Zparallel = 1 / Z1 + 1 / Z2
Zparallel

6
...
633633998940

Ztotal = Za + Zparallel
Ztotal (2dp)

HNC Electrical and Electronic Engineering

5
...
234439834025

Year Two: 2014/15

Keith A
...
Figure 1

𝑍 𝑎 = 2
...
5)

𝑍 𝑐 = 2
...
5)

e
𝑍 𝑏 = 1
...
25)

Figure 10: Circuit with Δ replaced by Y equivalent

The overall circuit impedance is shown in Table 9
...
Hudson

Electrical & Electronic Principles
06 November, 2014

17

4 Question 4
Series circuit: R-L-C
...
5 𝑘𝐻𝑧
𝑄𝑢𝑎𝑙𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑄 = 50

𝐶𝑖𝑟𝑐𝑢𝑖𝑡 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒,
𝑍 = 60 Ω (𝑎𝑡 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒)

𝐴𝑡 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒, 𝑋 𝐿 = 𝑋 𝐶
Because they are the same magnitude but opposite in sign they cancel out
...
190985931 𝐻
𝐿 = 190
...
000000021 𝐹
𝐶 = 21
...
22 𝑛𝐹 (2𝑑𝑝)

Year Two: 2014/15

Keith A
...
5 × 103 −

𝑁𝑜𝑡𝑒: 𝑓𝑟 𝑚𝑎𝑦 𝑎𝑙𝑠𝑜 𝑏𝑒 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒
𝑐𝑒𝑛𝑡𝑟𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦, 𝑓𝑐
...
5 × 10 − 25
𝑓𝑙 = 2475 𝐻𝑧
𝑓ℎ = 𝑓𝑟 +

2
...
5 × 103 +

50

∆𝑓 = 50 𝐻𝑧

50
2

3

𝑓ℎ = 2
...
985931 × 10−3
𝑋 𝐿 = 2970 Ω

𝑅 = 60 Ω

𝑋𝐶 =
𝑋𝐶 =

1
2𝜋𝑓 𝑙 𝐶
1
2×𝜋×2475×21
...
30303 Ω
𝑋 𝐶 = 3030 Ω (0dp)
𝑋 𝐿 − 𝑋 𝐶 = 2970 − 3030 = −60

∴ 𝑍 = (60 − 𝑗60) Ω

HNC Electrical and Electronic Engineering

Year Two: 2014/15

Keith A
...
985931 × 10−3
𝑋 𝐿 = 3030 Ω

𝑅 = 60 Ω

𝑋𝐶 =
𝑋𝐶 =

1
2𝜋𝑓ℎ 𝐶
1
2×𝜋×2525×21
...
29703 Ω
𝑋 𝐶 = 2970 Ω (0𝑑𝑝)
𝑋 𝐿 − 𝑋 𝐶 = 3030 − 2970 = 60

∴ 𝑍 = (60 + 𝑗60) Ω

HNC Electrical and Electronic Engineering

Year Two: 2014/15

Keith A
...
9999375 × 1010 )
141419
...
55622 𝐻𝑧
𝑓𝑟 = 22
...
Hudson

Electrical & Electronic Principles
06 November, 2014

21

Dynamic Resistance, Supply Current and Supply Voltage
At resonance, the current is in phase with the voltage
...
It is
called the Dynamic Resistance
...

𝐿

𝑅𝐷 =

𝐶𝑅

Figure 13: Dynamic Resistance Equation
Table 16: Dynamic resistance and inductor current calculations

𝐶 = 5 × 10−9 𝐹,

𝑅 = 25 Ω,

𝐼𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝐿 = 10 × 10−3 𝐻

𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑅 𝐷
10 × 10−3
𝑅𝐷 =
= 80,000 Ω = 80 kΩ
5 × 10−9 × 25
𝑆𝑢𝑝𝑝𝑙𝑦 𝐶𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼 𝑆 = 0
...

𝐼 𝑆 = 0
...
707 = 0
...
75 𝜇𝐴
𝐼 𝑆 = 176
...
75 × 10−6 × 80 × 103
𝑉 𝑆 = 14
...

Table 17: Capacitive reaction calculation

𝑓𝑟 = 22507
...
235667 Ω

2𝜋×22507
...
414 𝑘Ω (3𝑑𝑝)

0
...
Hudson

Electrical & Electronic Principles
06 November, 2014

22

Capacitor Branch
We can now calculate the current in the Capacitor branch of the circuit
...
14
1,414
...
009998333 A
= 9
...
00 𝑚𝐴 (𝑟𝑚𝑠)(2𝑑𝑝)
= (10
...

Table 19: Inductive reaction calculation

𝑓𝑟 = 22507
...
55622 × 10 × 10−3
= 1,414
...
414 𝑘Ω (3𝑑𝑝)

HNC Electrical and Electronic Engineering

𝐿 = 10 × 10−3 𝐻

Since we are at the resonant frequency
𝑋 𝐿 = 𝑋 𝐶 = 1
...
Hudson

Electrical & Electronic Principles
06 November, 2014

23

Current in the Inductor Branch

(Q5c)

We can now calculate the current in the Inductor branch of the circuit
...

Table 20: Current through the inductor

𝑅 = 25 Ω,

𝑋 𝐿 = 1,414
...
14 𝑉 (𝑟𝑚𝑠)

𝑇ℎ𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟 −
𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑏𝑟𝑎𝑛𝑐ℎ:
𝑉𝑆

1,414
...
522

𝐼 𝑅𝐿 =

𝑍 𝑅𝐿 = √2,000,687
...
456617 Ω
𝑍 𝑅𝐿 = 1
...
998333609 𝑚𝐴
𝐼 𝑅𝐿 = 10
...
00 ∠ − 90° 𝑚𝐴

𝑍 𝑅𝐿 =

√252

+

HNC Electrical and Electronic Engineering

𝑍 𝑅𝐿
14
...
235667

Year Two: 2014/15

Keith A
...
Hudson

25

Electrical & Electronic Principles
06 November, 2014

7 Question 7
Dot Notation

(Q7a)

When a transformer is wired up, the polarity of input and output connections often does not matter
...
However in applications where the output is linked in
some way to the input, the polarity does matter
...
When we talk of
polarity, we actually mean the direction of current flow at one instant in time
...
This is fine for an actual circuit but not for a
circuit diagram
...

If the (instantaneous) current enters one coil at the dotted terminal (positive), then the voltage induced in the
second coil will be positive at the dotted terminal
...
)

Figure 15: Dot notation (Potisuk, 2014)

Figure 15, Figure 16 (a) indicates the two coils that are wound in opposite directions, (b) indicates they are wound in
the same direction
...
Hudson

26

Electrical & Electronic Principles
06 November, 2014

(a)

(b)

Figure 16: Dot notation with respect to windings (Potisuk, 2014)

Figure 15 and Figure 16 shows two coils that are wound around a shared core
...


Figure 17: Series inductors (Potisuk, 2014)

HNC Electrical and Electronic Engineering

Year Two: 2014/15

Keith A
...
4

𝑀 = 𝑘√(𝐿1 𝐿2 ) = 0
...
0 = 𝐿1 + 𝐿2 + 2𝑀 ①

𝑆𝑒𝑟𝑖𝑒𝑠 − 𝑜𝑝𝑝𝑜𝑠𝑖𝑛𝑔:
𝐿 = 𝐿1 + 𝐿2 − 2𝑀, 𝐿 = 1
...
2 = 𝐿1 + 𝐿2 − 2𝑀 ②
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑀 𝑖𝑛𝑡𝑜 ① 𝑎𝑛𝑑 ②

2
...
4)√(𝐿1 𝐿2 )

1
...
4)√(𝐿1 𝐿2 )

2
...
8√(𝐿1 𝐿2 )

1
...
8√(𝐿1 𝐿2 )

③

③
+④
=

2
...
8√(𝐿1 𝐿2 )
1
...
8√(𝐿1 𝐿2 )
3
...
2
2

𝐷𝑖𝑣𝑖𝑑𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 2

④

=

2𝐿1 +2𝐿2

1
...
6 = 𝐿1 + 𝐿2
𝑅𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑒

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒
𝐿1 = 1
...
6 − 𝐿2
2
...
6 − 𝐿2 + 𝐿2 + 0
...
6 − 𝐿2 )𝐿2 )
2
...
6 + 0
...
6𝐿2 − 𝐿2 )
2
2
...
6 = 0
...
6𝐿2 − 𝐿2 )
2
0
...
8√(1
...
Hudson

Electrical & Electronic Principles
06 November, 2014

28
0
...
8

0
...
8√(1
...
8

0
...
6𝐿2 − 𝐿2 )
2
2

𝑆𝑞𝑢𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠

𝑀𝑜𝑣𝑒 𝑎𝑙𝑙 𝑡𝑜 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒

0
...
6𝐿2 − 𝐿2 )) 0
2
0
...
6𝐿2 − 𝐿2
2
𝐿2 − 1
...
25 = 0
2
+1𝐿2 − 1
...
25 = 0
2
𝐿2 =

𝑆𝑜𝑙𝑣𝑒 𝑡ℎ𝑒 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐, 𝑤ℎ𝑒𝑟𝑒
𝑎 = +1
𝑏 = −1
...
25

−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
2

𝐿2 =
𝐿2 =
𝐿2 =

−(−1
...
6) −4(1)(0
...
6±√(2
...
6±√1
...
4244998 𝑜𝑟 0
...
4244998
𝑖𝑛𝑡𝑜 𝐿1 = 1
...
6 − 1
...
1755111

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒
𝐿2 = 0
...
6 − 𝐿2

𝐿1 = 1
...
1755002
𝐿1 = 1
...
4244998 𝐻

HNC Electrical and Electronic Engineering

𝐿2 = 0
...
Hudson

29

Electrical & Electronic Principles
06 November, 2014

8 Bibliography
Nave, C
...
AC Thevenin's Theorem
...
phy-astr
...
edu/hbase/electric/acthev
...

Potisuk, S
...
, 2014
...
[Online]
Available at: http://faculty
...
edu/potisuk/elec202/notes/mcoupled
...


HNC Electrical and Electronic Engineering

Year Two: 2014/15



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