chapter 45 complex numbers AWS
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
757
© 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 4 758
© 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
759
© 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 4 760
© 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
762
© 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 764
© 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
766
© 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) 767
© 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° 768
© 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 = 6.986 769
© 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
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
757
© 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 4 758
© 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
759
© 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 4 760
© 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
762
© 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 764
© 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
766
© 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) 767
© 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° 768
© 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 = 6.986 769
© 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
