Download CHAPTER 45 COMPLEX NUMBERS

CHAPTER 45 COMPLEX NUMBERS
EXERCISE 187 Page 510
1. Solve the quadratic equation: x2 + 25 = 0
Since x 2 + 25 =
0 then
x 2 = −25
i.e.
x = −25 = (−1)(25) = −1 25 = j 25
from which,
x = ± j5
2. Solve the quadratic equation: x2 – 2x + 2 = 0
Since x2 – 2x + 2 = 0 then
=
x
−−2±
[(−2)2 − 4(1)(2)]
2(1)
=
2 ± −4 2 ± (−1)(4) 2 ± (−1) (4) 2 ± j (4)
=
=
=
2
2
2
2
=
2
4 2
2
±j
=± j = 2 ± j1 = 1 ± j
2
2
2
2
3. Solve the quadratic equation: x2 – 4x + 5 = 0
Since x2 – 4x + 5 = 0 then
=
x
−−4±
[(−4)2 − 4(1)(5)]
2(1)
=
4 ± −4 4 ± (−1)(4) 4 ± (−1) (4) 4 ± j (4)
=
=
=
2
2
2
2
=
4
4 4
2
±j
=± j = 2 ± j1 = 2 ± j
2
2
2
2
4. Solve the quadratic equation: x2 – 6x + 10 = 0
Since x2 – 6x + 10 = 0 then
=
x
−−6±
[(−6)2 − 4(1)(10)]
2(1)
=
6 ± −4 6 ± (−1)(4) 6 ± (−1) (4) 6 ± j (4)
=
=
=
2
2
2
2
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© 2014, John Bird
=
6
4 6
2
±j
=± j = 3 ± j
2
2
2
2
5. Solve the quadratic equation: 2x2 – 2x + 1 = 0
Since 2x2 – 2x + 1 = 0 then
=
x
−−2±
[(−2)2 − 4(2)(1)]
=
2(2)
2 ± −4 2 ± (−1)(4) 2 ± (−1) (4) 2 ± j (4)
=
=
=
4
4
4
4
=
2
4 2
2
±j
=± j = 0.5 ± j0.5
4
4
4
4
6. Solve the quadratic equation: x2 – 4x + 8 = 0
Since x2 – 4x + 8 = 0 then
x
=
−−4±
[(−4)2 − 4(1)(8)]
2(1)
=
4 ± −16 4 ± (−1)(16) 4 ± (−1) (16) 4 ± j (16)
=
=
=
2
2
2
2
=
4
16 4
4
±j
=
± j = 2 ± j2
2
2
2
2
7. Solve the quadratic equation: 25x2 – 10x + 2 = 0
Since 25x2 – 10x + 2 = 0 then
=
x
− − 10 ±
[(−10)2 − 4(25)(2)]
2(25)
=
=
10 ± −100 10 ± (−1)(100) 10 ± (−1) (100)
=
=
50
50
50
10 ± j (100)
10
100 10
10
=
= 0.2 ± j0.2
±j
=± j
50
50
50
50
50
8. Solve the quadratic equation: 2x2 + 3x + 4 = 0
Since 2 x 2 + 3 x + 4 =
0 then
=
x
−3 ±
[32 − 4(2)(4)]
=
2(2)
−3 ± −23 −3 ± (−1)(23) −3 ± (−1) (23) −3 ± j (23)
=
=
=
4
4
4
4
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© 2014, John Bird
3
23
=– ± j
4
4
or (– 0.750 ± j1.199)
9. Solve the quadratic equation: 4t2 – 5t + 7 = 0
Since 4t2 – 5t + 7 = 0 then
=
t
[(−5)2 − 4(4)(7)]
−−5±
=
2(4)
5 ± −87 5 ± (−1)(87) 5 ± (−1) (87) 5 ± j (87)
=
=
=
8
8
8
8
=
10. Evaluate (a) j8 (b) –
1
j7
(c)
5
87
±j
8
8
or (0.625 ± j1.166)
4
2 j13
(a) j 8 = ( j 2 ) = ( −1) = 1
4
4
(b) j 7 = j × j 6 = j × ( j 2 ) = j × ( −1) = − j
3
Hence,
−
3
−j
−j
−j
−j
1
1 1
=
−
==
= =
= = –j
7
2
− j j j (− j ) − j
−(−1) 1
j
(c) j13 = j × j12 = j × ( j 2 ) = j × (−1)6 = j
6
Hence,
4
2 2(− j ) − j 2 − j 2
= =
==
= –j2
j j (− j ) − j 2
1
2 j13
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© 2014, John Bird
EXERCISE 188 Page 513
1. Evaluate (a) (3 + j2) + (5 – j) and (b) (–2 + j6) – (3 – j2) and show the results on an Argand diagram.
(a) (3 + j2) + (5 – j) = (3 + 5) + j(2 – 1) = 8 + j
(b) (–2 + j6) – (3 – j2) = –2 + j6 – 3 + j2 = (–2 – 3) + j(6 + 2) = –5 + j8
(8 + j) and (–5 + j8) are shown on the Argand diagram below.
2. Write down the complex conjugates of (a) 3 + j4, (b) 2 – j.
(a) The complex conjugate of 3 + j4 is: 3 – j4
(b) The complex conjugate of 2 – j is: 2 + j
3. If z = 2 + j and w = 3 – j evaluate (a) z + w
(b) w – z
(c) 3z – 2w
(d) 5z + 2w (e) j(2w – 3z) (f) 2jw – jz
(a) z + w = (2 + j) + (3 – j) = 2 + j + 3 – j = 5
(b) w – z = (3 – j) – (2 + j) = 3 – j – 2 – j = 1 – j2
(c) 3z – 2w = 3(2 + j) – 2(3 – j) = 6 + j3 – 6 + j2 = j5
(d) 5z + 2w = 5(2 + j) + 2(3 – j) = 10 + 5j + 6 – j2 = 16 + j3
(e) j(2w – 3z) = j[(6 – j2) – (6 +j3)] = j[6 – j2 – 6 – j3] = j(– j5) = – j 2 5 = –(– 1)5 = 5
(f) 2jw – jz = 2j(3 – j) – j(2 + j) = j6 – 2 j 2 – j2 – j 2 = j6 – 2(– 1) – j2 – (– 1) = j6 + 2 – j2 + 1 = 3 +
j4
4. Evaluate in a + jb form, given Z 1 = 1 + j2, Z2 = 4 – j3, Z 3 = –2 + j3 and Z4 = –5 – j
(a) Z 1 + Z 2 – Z 3
(b) Z 2 – Z1 + Z 4
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© 2014, John Bird
(a) Z1 + Z 2 − Z 3 = 1 + j2 + 4 – j3 – (–2 + j3) = 1 + j2 + 4 – j3 + 2 – j3
= (1 + 4 + 2) + j(2 – 3 – 3) = 7 – j4
(b) Z 2 − Z1 + Z 4 = (4 – j3) – (1 + j2) + (–5 – j) = 4 – j3 – 1 – j2 – 5 – j
= (4 – 1 – 5) + j(–3 – 2 – 1) = – 2 – j6
5. Evaluate in a + jb form, given Z 1 = 1 + j2, Z2 = 4 – j3, Z 3 = – 2 + j3 and Z4 = –5 – j
(a) Z 1 Z 2
(b) Z 3 Z 4
(a) Z 1 Z 2 = (1 + j2)(4 – j3) = 4 – j3 + j8 – j 2 6 = 4 – j3 + j8 + 6 = 10 + j5
(b) Z3 Z 4 = (–2 + j3)(–5 – j) = 10 + j2 – j15 – j 2 3 = 10 + j2 – j15 + 3 = 13 – j13
6. Evaluate in a + jb form, given Z 1 = 1 + j2, Z2 = 4 – j3, Z 3 = –2 + j3 and Z4 = –5 – j
(a) Z 1 Z 3 + Z 4
(b) Z 1 Z 2 Z3
(a) Z1Z 3 + Z 4 = (1 + j2)(–2 + j3) + (–5 – j) = –2 + j3 – j4 + j 2 6 – 5 – j
= –2 + j3 – j4 – 6 – 5 – j = –13 – j2
(b) Z1Z 2 Z 3 = (1 + j2)(4 – j3)(–2 + j3) = (4 – j3 + j8 – j 2 6)(–2 + j3) = ( 10 + j5)(–2 + j3)
= –20 + j30 – j10 + j 2 15 = –20 + j30 –j10 –15 = –35 + j20
7. Evaluate in a + jb form, given Z 1 = 1 + j2, Z2 = 4 – j3, Z 3 = –2 + j3 and Z4 = –5 – j
(a)
Z1
Z2
(b)
Z1 + Z 3
Z2 − Z4
Z1 1 + j 2 (1 + j 2)(4 + j 3) 4 + j 3 + j8 + j 2 6 4 + j 3 + j8 − 6 −2 + j11 − 2
11
=
=
=
(a)= =
=
+j
2
2
Z 2 4 − j 3 (4 − j 3)(4 + j 3)
4 +3
25
25
25
25
Z1 + Z 3 (1 + j 2) + (−2 + j 3) −1 + j 5 (−1 + j 5)(9 + j 2) −9 − j 2 + j 45 + j 210 −19 + j 43
= =
=
=
(b) =
Z 2 − Z 4 (4 − j 3) − (−5 − j )
9 − j2
(9 − j 2)(9 + j 2)
92 + 22
85
=
761
−19
43
+j
85
85
© 2014, John Bird
8. Evaluate in a + jb form, given Z 1 = 1 + j2, Z2 = 4 – j3, Z 3 = –2 + j3 and Z4 = –5 – j
(a)
(a)
Z1Z 3
Z1 + Z 3
(b) Z 2 +
Z1
+ Z3
Z4
(1 + j 2)(−2 + j 3)
−2 + j 3 − j 4 + j 2 6 −8 − j
Z1Z 3
= = =
(1 + j 2) + (−2 + j 3)
−1 + j 5
−1 + j 5
Z1 + Z 3
=
(b) Z 2 +
(−8 − j )(−1 − j 5) 8 + j 40 + j + j 2 5 3 + j 41
3
41
= =
=
+j
2
2
(−1 + j 5)(−1 − j 5)
1 +5
26
26
26
1+ j2
Z1
(1 + j 2)(−5 + j )
+ (–2 + j3) = 4 – j3 +
– 2 + j3
+ Z 3 = (4 – j3) +
−5 − j
Z4
52 + 12
= 4 – j3 +
−5 + j − j10 + j 2 2
– 2 + j3
26
= 4 – j3 +
−7 − j 9
7
9
– 2 + j3 = 2 –
−j
26
26
26
=
9. Evaluate (a)
1− j
1+ j
(b)
52 7
9
45
9
=
− −j
−j
26 26
26
26
26
1
1+ j
1 − j (1 − j )(1 − j ) 1 − j − j + j 2 − j 2
=
= =
(a)
= 0 – j1 = – j
1 + j (1 + j )(1 − j )
12 + 12
2
1
(1)(1 − j )
1− j 1− j
1
1
= =
(b) =
= −j
2
2
1 + j (1 + j )(1 − j ) 1 + 1
2
2
2
10. Show that
−25  1 + j 2 2 − j 5 
−

 = 57 + j24
−j 
2  3 + j4
1 + j 2 (1 + j 2)(3 − j 4) 3 − j 4 + j 6 − j 2 8 11 + j 2 11
2
=
=
=
=
+j
2
2
3 + j4
3 +4
25
25
25
25
2 − j 5 (2 − j 5)( j )
=
=
−j
− j( j)
j2 − j2 5 5 + j2
=
= 5 + j2
− j2
1
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© 2014, John Bird
L.H.S. =

−25  1 + j 2 2 − j 5 
25  11
2 
25  11

−
−  + j  − (5 + j 2)  =
−  − 5  +

=
−j 
2  3 + j4
2  25
25 
2  25 

=−
25  11 − 125 

+
2  25 
=−
25  114 
25  48 
−
 + j   = 57 + j24 = R.H.S.
2  25 
2  25 
763
 2

j  − 2 
 25

25  114
48 
 2 − 50  
− −
−j 
j
 =
2  25
25 
 25  
© 2014, John Bird
EXERCISE 189 Page 514
1. Solve: (2 + j)(3 – j2) = a + jb
(2 + j)(3 – j2) = a + jb
Hence,
6 – j4 + j3 – j 2 2 = a + jb
i.e.
Thus,
8 – j1 = a + jb
a = 8 and b = –1
2. Solve:
2+ j
= j(x + jy)
1− j
2+ j
(2 + j )(1 + j )
= j ( x + jy ) hence,
= j ( x + jy )
1− j
(1 − j )(1 + j )
i.e.
2 + j2 + j + j2
= jx + j 2 y
12 + 12
i.e.
1 + j3
= jx − y
2
i.e.
1
3
+ j = –y + jx
2
2
Hence,
3. Solve: (2 – j3) =
x=
3
2
y= −
and
1
2
(a + b)
(2 − j 3) =
Squaring both sides gives:
( 2 − j 3)
2
(a + jb)
=
a + jb
(2 – j3)(2 – j3) = a + jb
i.e.
4 – j6 – j6 + j 2 9 = a + jb
i.e.
–5 – j12 = a + jb
Hence,
a = –5
and
b = –12
4. Solve: (x – j2y) – (y – jx) = 2 + j
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© 2014, John Bird
(x – j2y) – (y – jx) = 2 + j
Hence,
(x – y) + j(– 2y + x) = 2 + j
i.e.
x–y=2
(1)
and
x – 2y = 1
(2)
(1) – (2) gives:
y=1
Substituting in (1) gives: x – 1 = 2 from which, x = 3
5. If Z = R + jωL + 1/jωC, express Z in (a + jb) form when R = 10, L = 5, C = 0.04 and ω = 4
Z = R + jωL +
1
6.25
6.25(− j )
1
= 10 + j(4)(5) +
= 10 + j20 +
= 10 + j20 +
jωC
j
j (− j )
j (4)(0.04)
= 10 + j20 –
6.25
= 10 + j20 – j6.25
− j2
= 10 + j13.75
765
© 2014, John Bird
EXERCISE 190 Page 517
1. Determine the modulus and argument of (a) 2 + j4 (b) –5 – j2 (c) j(2 – j)
(a) 2 + j4 lies in the first quadrant as shown below
Modulus, r =
42 + 22 = 4.472
Argument, θ = tan −1
4
= 63.43°
2
(b) –5 – j2 lies in the third quadrant as shown below
Modulus, r =
α = tan −1
52 + 22 = 5.385
2
= 21.80°
5
Hence, argument, θ = –(180° – 21.80°) = –158.20°
(c) j(2 – j) = j2 – j 2 = j2 + 1 or 1 + j2
1 + j2 lies in the first quadrant as shown below.
Modulus, r = 12 + 22 = 2.236
Argument, θ = tan −1
2
= 63.43°
1
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© 2014, John Bird
2. Express in polar form, leaving answers in surd form:
(a) 2 + j3
(b) –4
(c) –6 + j
(a) 2 + j3 From the diagram below, r =
22 + 32 =
13
3
 
θ =tan −1   =56.31° or 56°19 '
2
and
Hence, 2 + j3 = 13∠56.31° in polar form
(b) –4 = –4 + j0 and is shown in the diagram below, where r = 4 and θ = 180°
Hence,
(c) –6 + j
and
Thus,
–4 = 4∠180° in polar form
From the diagram below, r =
1
=
α tan −1  =
 9.46°
6
–6 + j =
thus
62 + 12 =37
θ = 180° – 9.46° = 170.54°
37∠170.54°
3. Express in polar form, leaving answers in surd form:
(a) – j3
(b) (– 2 + j)3
(c) j3(1 – j)
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© 2014, John Bird
(a) –j3
From the diagram below, r = 3 and θ = –90°
Hence,
(b)
( −2 + j )
– j3 = 3∠–90° in polar form
3
= (–2 + j)(–2 + j)(–2 + j) = (4 – j2 – j2 + j 2 )(–2 + j)
= (3 – j4)(–2 + j) = –6 + j3 + j8 – j 2 4 = –2 + j11
From the diagram below, r =
θ = 180° – 79.70° = 100.30°
and
Hence,
 11 
and =
α tan −1 =
 79.70°
2
22 + 112 =
125
( −2 + j )
3
= –2 + j11 = 125∠100.30° in polar form
(c) j 3 (1 − j ) = (j)( j 2 )(1 – j) = –j(1 – j) = –j + j 2 = –1 – j
From the diagram below, r = 12 + 12 =2
1
α= tan −1  =
 45°
1
θ = 180° – 45° = 135°
and
Hence,
and
j 3 (1 − j ) = –1 – j =
2∠ − 135°
4. Convert into (a + jb) form giving answers correct to 4 significant figures:
(a) 5∠30°
(b) 3∠60°
(c) 7∠45°
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© 2014, John Bird
(a) 5∠30° = 5 cos 30° + j 5 sin 30° = 4.330 + j2.500
(b) 3∠60° = 3 cos 60° + j 3 sin 60° = 1.500 + j2.598
(c) 7∠45° = 7 cos 45° + j 7 sin 45° = 4.950 + j4.950
5. Convert into (a + jb) form giving answers correct to 4 significant figures:
(a) 6∠125° (b) 4∠π (c) 3.5∠–120°
(a) 6∠125° = 6 cos 125° + j 6 sin 125° = –3.441 + j4.915
(b) 4∠π = 4 cos π + j sin π = – 4.000 + j0
(Note that π is radians)
(c) 3.5∠–120° = 3.5 cos(–120°) + j 3.5 sin(–120°) = –1.750 – j3.031
6. Evaluate in polar form: (a) 3∠20° × 15∠45°
(b) 2.4∠65° × 4.4∠–21°
(a) 3∠20°×15∠45° = 3 ×15∠(20° + 45°) = 45∠65°
(b) 2.4∠65°× 4.4∠ − 21° = 2.4 × 4.4∠(65° + −21°) = 10.56∠44°
7. Evaluate in polar form: (a) 6.4∠27° ÷ 2∠–15° (b) 5∠30° × 4∠80° ÷ 10∠–40°
(a) 6.4∠27° ÷ 2∠ − 15° =
6.4∠27° 6.4
=
∠27° − −15° = 3.2∠42°
2∠ − 15°
2
(b) 5∠30°× 4∠80° ÷ 10∠ − 40° =
5∠30°× 4∠80° 5 × 4
=
∠(30° + 80° − −40°) = 2∠150°
10∠ − 40°
10
8. Evaluate in polar form: (a) 4 ∠
(a) 4∠
π
6
+ 3∠
π
6
+ 3∠
π
8
(b) 2∠120° + 5.2∠58° – 1.6∠– 40°
π
π 
π
π

=  4 cos + j 4sin  +  3cos + j 3sin  = (3.464 + j2) + (2.772 + j1.148)
6
6 
8
8
8

π
= 6.236 + j3.148
From the diagram below, r =
6.2362 + 3.1482 =
6.986
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© 2014, John Bird
 3.148 
=
θ tan −1  =
 26.79° or 0.467 rad
 6.236 
and
Hence,
4∠
π
6
+ 3∠
π
8
= 6.986∠26.79°
or
6.986∠0.467 rad
(b) 2∠120° + 5.2∠58° − 1.6∠ − 40°
= (2 cos 120° + j2 sin 120°) + (5.2 cos 58° + j5.2 sin 58°) – (1.6 cos(–40°) + j1.6 sin(–40°))
= (–1 + j1.732) + (2.756 + j4.410) – (1.226 – j1.028)
= –1 + j1.732 + 2.756 + j4.410 – 1.226 + j1.028
= 0.530 + j7.170
From the diagram below, r =
and
Hence,
0.5302 + 7.1702 =
7.190
 7.170 
θ tan −1  =
=
 85.77°
 0.530 
2∠120° + 5.2∠58° − 1.6∠ − 40° = 7.190∠85.77°
770
© 2014, John Bird
EXERCISE 191 Page 519
1. Determine the resistance R and series inductance L (or capacitance C) for each of the following
impedances assuming the frequency to be 50 Hz:
(a) (3 + j8) Ω
(b) (2 – j3) Ω
(c) j14 Ω
(d) 8∠– 60° Ω
(a) If Z = (3 + j8) Ω then resistance, R = 3 Ω
and inductive reactance, X L = 8 Ω (since the j
term is positive)
X L = 2πfL = 8 hence, inductance, L =
8
2π f
(b) If Z = (2 – j3) Ω then resistance, R = 2 Ω
=
8
= 0.0255 H or 25.5 mH
2π (50)
and capacitive reactance, X C = 3 Ω (since the j
term is negative)
XC =
1
1
1
1.061× 10−3 or 1061× 10−6
= =
= 3 hence, capacitance, C =
2π f (3) 2π (50)(3)
2π fC
= 1061 µF
(c) If Z = j14 Ω i.e. Z = (0 + j14) Ω then resistance, R = 0 Ω
i.e.
2πfL = 14 hence, inductance, L =
and X L = 14 Ω
14
= 0.04456 H or 44.56 mH
2π (50)
(d) If Z = 8∠ − 60°Ω = 8 cos(–60°) + j8 sin(–60°) = (4 – j6.928) Ω
Hence,
i.e.
resistance, R = 4 Ω
1
= 6.928
2π fC
and X C = 6.928 Ω
and capacitance, C =
1
=
2π (50)(6.928)
459.4 ×10−6 = 459.4 µF
2. Two impedances, Z1 = (3 + j6) Ω and Z 2 = (4 – j3) Ω are connected in series to a supply voltage of
120 V. Determine the magnitude of the current and its phase angle relative to the voltage
In a series circuit, total impedance, ZTOTAL= Z1 + Z 2 = (3 + j6) + (4 – j3) = (7 + j3) Ω
=
3
7 2 + 32 ∠ tan −1  
7
= 7.616∠23.20° Ω
Since voltage V = 120∠0° V, then current, I =
V
120∠0°
= 15.76∠– 23.20° A
=
Z 7.616∠23.20°
i.e. the current is 15.76 A and is lagging the voltage by 23.20°
771
© 2014, John Bird
3. If the two impedances in Problem 2 are connected in parallel, determine the current flowing and its
phase relative to the 120 V supply voltage.
In a parallel circuit shown below, the total impedance ZT is given by:
1
1
1
1
1
3 − j 6 4 + j3
3
6
4
3
= +
=
+
=
+
=
−j + +j
ZT Z1 Z 2 3 + j 6 4 − j 3 32 + 62 42 + 32 45
45 25
25
1
= admittance, YT = 0.22667 – j0.01333 = 0.2271∠–3.37° siemen
ZT
i.e.
V
Current, I = = VY
=
°) 27.25∠ − 3.37° A
(120∠0°)(0.2271∠ − 3.37=
T
ZT
i.e. the current is 27.25 A and is lagging the voltage by 3.37°
4. A series circuit consists of a 12 Ω resistor, a coil of inductance 0.10 H and a capacitance of 160 µF.
Calculate the current flowing and its phase relative to the supply voltage of 240 V, 50 Hz. Determine
also the power factor of the circuit.
R = 12 Ω , inductive reactance, X L = 2πfL = 2π(50)(0.10) = 31.416 Ω
and capacitive reactance,
=
XC
1
1
= 19.894 Ω
=
2π f C 2π ( 50 )(160 ×10−6 )
Hence, impedance, Z = R + j( X L − X C ) = 12 + j(31.416 – 19.894) = (12 + j11.52) Ω = 16.64∠43.83° Ω
Current flowing, I =
240∠0°
V
= 14.42∠– 43.83° A
=
Z 16.64∠43.83°
Phase angle = 43.83° lagging (i.e. I lags V by 43.83°)
Power factor = cos ϕ = cos 43.83° = 0.721
772
© 2014, John Bird
5. For the circuit shown, determine the current I flowing and its phase relative to the applied voltage.
1
1
1
1
1
1
1
30 + j 20 40 − j 50 1
= +
+
=
+
+
=
+
+
ZT Z1 Z 2 Z 3 30 − j 20 40 + j 50 25 302 + 202 402 + 502 25
=
30
20
40
50
1
+j
+
−j
+
1300
1300 4100
4100 25
1
= admittance,YT = 0.07283 + j0.00319 = 0.0729∠2.51° S
ZT
i.e.
Current, I =
V
= VYT = (200∠0°)(0.0729∠2.51°) = 14.6∠2.51° A
ZT
i.e. the current is 14.6 A and is leading the voltage by 2.51°
6. Determine, using complex numbers, the magnitude and direction of the resultant of the coplanar forces
given below, which are acting at a point. Force A, 5 N acting horizontally, Force B, 9 N acting at an
angle of 135° to force A, Force C, 12 N acting at an angle of 240° to Force A.
Resultant force = FA + FB + FC = 5∠0° + 9∠135° + 12∠240°
= (5 + j0) + (–6.364 + j6.364) + (–6 – j10.392)
= 5 + j0 – 6.364 + j6.364 – 6 – j10.392)
= –7.364 – j4.028
= 8.394∠– 151.32° or 8.394∠208.68° N
Hence, the magnitude of the force that has the same effect as the three forces acting separately is:
8.392 N and its direction is 208.68° to the horizontal (i.e. from Force A)
773
© 2014, John Bird
7. A delta-connected impedance Z A is given by:
ZA =
Z1Z 2 + Z 2 Z 3 + Z 3 Z1
Z2
Determine Z A in both Cartesian and polar form given Z 1 = (10 + j0) Ω, Z2 = (0 – j10) Ω and
Z 3 = (10 + j10) Ω
ZA
Z1Z 2 + Z 2 Z 3 + Z 3 Z1 (10 + j 0)(0 − j10) + (0 − j10)(10 + j10) + (10 + j10)(10 + j 0)
=
Z2
(0 − j10)
=
− j100 − j100 − j 2100 + 100 + j100 200 − j100 200
j100
=
=
−
− j10
− j10
− j10 − j10
=
j 200
+ 10 = (10 + j 20) Ω
10
From the diagram below, r = 102 + 202 = 22.36
Hence,
and
 20 
=
θ tan −1  =
 63.43°
 10 
(10 + j 20) Ω
Z=
= 22.36∠63.43°Ω
A
8. In the hydrogen atom, the angular momentum p of the de Broglie wave is given by
 jh 
pψ = – 
 (± jmψ). Determine an expression for p.
 2π 
If
h 2
 jh
 j h  ± j m Ψ 
 jh
−
−
pΨ = − 
( j )( ± m )
 ( ± j m Ψ ) then p = − 

=
 (± j m) =
2π
 2π 
 2π  Ψ 
 2π 
=
774
mh
h
( ±m ) = ±
2π
2π
© 2014, John Bird
9. An aircraft P flying at a constant height has a velocity of (400 + j300)km/h. Another aircraft Q at the
same height has a velocity of (200 – j600) km/h. Determine (a) the velocity of P relative to Q, and
(b) the velocity of Q relative to P. Express the answers in polar form, correct to the nearest km/h.
(a) The velocity of P relative to Q = vP − vQ = (400 + j300) – (200 – j600)
= 400 + j300 – 200 + j600
= 200 + j900 = 922∠77.47°
i.e. the velocity of P relative to Q is 922 km/h at 77.47°
(b) The velocity of Q relative to P = vQ − vP = (200 – j600) – (400 + j300)
= 200 – j600 – 400 – j300
= –200 – j900 = 922∠–102.53°
i.e. the velocity of Q relative to P is 922 km/h at –102.53°
10. Three vectors are represented by P, 2∠30°, Q, 3∠90° and R, 4∠–60°. Determine in polar form the
vectors represented by (a) P + Q + R, (b) P – Q – R.
(a) P + Q + R = 2∠30° + 3∠90° + 4∠–60° = (1.732 + j1) + (0 + j3) + (2 – j3.464)
= (3.732 + j0.536) = 3.770∠8.17°
(b) P – Q – R = 2∠30° – 3∠90° – 4∠–60° = (1.732 + j1) – (0 + j3) – (2 – j3.464)
= (–0.268 + j1.464)
 1.464 
α tan −1  =
From the diagram below, r = 1.488 and =
 79.63°
 0.268 
and
=
θ 180° − 79.63=
° 100.37°
775
© 2014, John Bird
Hence,
P – Q – R = 1.488∠100.37°
11. In a Schering bridge circuit, =
Zx
( RX − jX C ) ,
X
Z 2 = − jX C2 , Z 3 =
( R3 )( − jX C )
( R3 − jX C )
3
and
3
Z 4 = R4 where X C =
1
. At balance: ( Z X )( Z 3 ) = ( Z 2 )( Z 4 ) .
2π fC
Show that at balance RX =
( Z X )( Z3 ) = ( Z 2 )( Z 4 )
Since
then
C3 R4
C2 R3
and C X =
R4
C2
 ( R3 )( − jX C3 ) 
( RX − jX C X ) 
( − jX C2 )( R4 )
=
R
−
jX
3
C
3


Thus,
( R3 − jX C3 )( − jX C2 )( R4 )
( RX − jX C X ) =
( R3 )( − jX C3 )
i.e.
( RX − jX
=
CX )
− j R3 X C2 R4
j 2 X C3 X C2 R4
+
( R3 )( − jX C3 ) ( R3 )( − jX C3 )
i.e.
( RX − jX C X ) =
X C2 R4
X C2 R4
−
X C3
( R3 ) (− j )
=
i.e.
Equating the real parts gives:
i.e.
( RX − jX C X ) =
X C2 R4 X C2 R4
+
X C3
j R3
X C2 R4
X C R4
−j 2
X C3
R3
1
R4
X C2 R4 2π fC2
2π fC3
RX
R4
=
=
=
1
X C3
2π fC2
2π fC3
RX =
C3 R4
C2
X C R4
Equating the imaginary parts gives: − X C X =
− 2
R3
i.e.
from which,
1
R4
1
R4
2π fC2
= =
2π fC X
2π fC2 R3
R3
CX =
C2 R3
R4
776
© 2014, John Bird
12. An amplifier has a transfer function T given by T =
500
where ω is the angular
1 + jω ( 5 ×10−4 )
frequency. The gain of the amplifier is given by the modulus of T and the phase is given by the
argument of T. If ω = 2000 rad/s, determine the gain and the phase (in degrees).
500
500
500
When ω = 2000 rad/s , transfer function T=
=
=
−
4
−
4
1 + jω ( 5 ×10 ) 1 + j (2000)(5 ×10 ) 1 + j1
500
(500)(1 − j1) 500 − j 500 500 − j 500 500
500
= = =
−j
Hence, =
T=
1 + j1 (1 + j1)(1 − j1)
12 + 12
2
2
2
= 250 – j250 = 353.6∠– 45°
Hence, the gain of the amplifier = 353.6 and the phase is –45°
13. The sending end current of a transmission line is given by I S =
VS
tanh PL . Calculate the
Z0
value of the sending current, in polar form, given VS = 200 V , Z 0 =560 + j 420 Ω , P = 0.20
and L = 10
=
IS
Sending current,
VS
200
200 tanh 2
tanh
=
PL
tanh ( 0.20=
×10 )
Z0
(560 + j 420)
(560 + j 420)
192.8
(192.8)(560 − j 420)
(192.8)(560 − j 420)
=
=
=
(560 + j 420) (560 + j 420)(560 − j 420)
5602 + 4202
=
(192.8) ( 700∠ − 36.87° )
= 0.275∠ − 36.87° A
490000
i.e. the sending end current, I S = 275∠ − 36.87° mA
777
© 2014, John Bird