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© 2014, John Bird

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CHAPTER 45 COMPLEX NUMBERS

EXERCISE 187 Page 510

1. Solve the quadratic equation: x2 + 25 = 0

Since 2 25 0x + = then 2 25x = − i.e. 25 ( 1)(25) 1 25 25x j= − = − = − = from which, x = ± j5 2. Solve the quadratic equation: x2 – 2x + 2 = 0

Since x2 – 2x + 2 = 0 then

[ ]22 ( 2) 4(1)(2) 2 ( 1)(4) 2 ( 1) (4) 2 (4)2 4

2(1) 2 2 2 2j

x− − ± − − ± − ± − ±± −

= = = = =

= 2 4 2 22 2 2 2

j j± = ± = 2 ± j1 = 1 ± j

3. Solve the quadratic equation: x2 – 4x + 5 = 0


[ ]24 ( 4) 4(1)(5) 4 ( 1)(4) 4 ( 1) (4) 4 (4)4 4

2(1) 2 2 2 2j

x− − ± − − ± − ± − ±± −

= = = = =

= 4 4 4 22 2 2 2

j j± = ± = 2 ± j1 = 2 ± j



[ ]26 ( 6) 4(1)(10) 6 ( 1)(4) 6 ( 1) (4) 6 (4)6 4

2(1) 2 2 2 2j

x− − ± − − ± − ± − ±± −

= = = = =

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= 6 4 6 22 2 2 2

j j± = ± = 3 ± j

5. Solve the quadratic equation: 2x2 – 2x + 1 = 0

Since 2x2 – 2x + 1 = 0 then

[ ]22 ( 2) 4(2)(1) 2 ( 1)(4) 2 ( 1) (4) 2 (4)2 4

2(2) 4 4 4 4j

x− − ± − − ± − ± − ±± −

= = = = =

= 2 4 2 24 4 4 4

j j± = ± = 0.5 ± j0.5



[ ]24 ( 4) 4(1)(8) 4 ( 1)(16) 4 ( 1) (16) 4 (16)4 16

2(1) 2 2 2 2j

x− − ± − − ± − ± − ±± −

= = = = =

= 4 16 4 42 2 2 2

j j± = ± = 2 ± j2

7. Solve the quadratic equation: 25x2 – 10x + 2 = 0

Since 25x2 – 10x + 2 = 0 then

[ ]210 ( 10) 4(25)(2) 10 ( 1)(100) 10 ( 1) (100)10 100

2(25) 50 50 50x

− − ± − − ± − ± −± −= = = =

= 10 (100)

50j±

= 10 100 10 1050 50 50 50

j j± = ± = 0.2 ± j0.2

8. Solve the quadratic equation: 2x2 + 3x + 4 = 0

Since 22 3 4 0x x+ + = then

[ ]23 3 4(2)(4) 3 ( 1)(23) 3 ( 1) (23) 3 (23)3 23

2(2) 4 4 4 4j

x− ± − − ± − − ± − − ±− ± −

= = = = =

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= – 3 234 4

j± or (– 0.750 ± j1.199)

9. Solve the quadratic equation: 4t2 – 5t + 7 = 0

Since 4t2 – 5t + 7 = 0 then

[ ]25 ( 5) 4(4)(7) 5 ( 1)(87) 5 ( 1) (87) 5 (87)5 87

2(4) 8 8 8 8j

t− − ± − − ± − ± − ±± −

= = = = =

= 5 878 8

j± or (0.625 ± j1.166)

10. Evaluate (a) j8 (b) –7

1j

(c) 13

42 j

(a) 8j = ( )42j = ( )41− = 1

(b) ( ) ( )3 37 6 2 1j j j j j j j= × = × = × − = −

Hence, 7 2

1 1 1( ) ( 1) 1

j j j jj j j j j j

− − − −− = − = = = = =

− − − − − = –j

(c) ( )613 12 2 6( 1)j j j j j j j= × = × = × − =

Hence, 13

42 j

= 2

2 2( ) 2 2( ) 1

j j jj j j j

− − −= = =

− − = –j2

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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) = – 2j 5 = –(– 1)5 = 5 (f) 2jw – jz = 2j(3 – j) – j(2 + j) = j6 – 2 2j – j2 – 2j = j6 – 2(– 1) – j2 – (– 1) = j6 + 2 – j2 + 1 = 3 + j4 4. Evaluate in a + jb form, given Z1 = 1 + j2, Z2 = 4 – j3, Z3 = –2 + j3 and Z4 = –5 – j

(a) Z1 + Z2 – Z3 (b) Z2 – Z1 + Z4

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(a) 1 2 3Z Z Z+ − = 1 + j2 + 4 – j3 – (–2 + j3) = 1 + j2 + 4 – j3 + 2 – j3 = (1 + 4 + 2) + j(2 – 3 – 3) = 7 – j4 (b) 2 1 4Z Z Z− + = (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 Z1 = 1 + j2, Z2 = 4 – j3, Z3 = – 2 + j3 and Z4 = –5 – j

(a) Z1 Z2 (b) Z3 Z4

(a) Z1 Z2 = (1 + j2)(4 – j3) = 4 – j3 + j8 – 2 6j = 4 – j3 + j8 + 6 = 10 + j5 (b) Z3 Z4 = (–2 + j3)(–5 – j) = 10 + j2 – j15 – 23j = 10 + j2 – j15 + 3 = 13 – j13 6. Evaluate in a + jb form, given Z1 = 1 + j2, Z2 = 4 – j3, Z3 = –2 + j3 and Z4 = –5 – j

(a) Z1 Z3 + Z4 (b) Z1 Z2 Z3

(a) 1 3 4Z Z Z+ = (1 + j2)(–2 + j3) + (–5 – j) = –2 + j3 – j4 + 2j 6 – 5 – j

= –2 + j3 – j4 – 6 – 5 – j = –13 – j2 (b) 1 2 3Z Z Z = (1 + j2)(4 – j3)(–2 + j3) = (4 – j3 + j8 – 2j 6)(–2 + j3) = ( 10 + j5)(–2 + j3)

= –20 + j30 – j10 + 2j 15 = –20 + j30 –j10 –15 = –35 + j20 7. Evaluate in a + jb form, given Z1 = 1 + j2, Z2 = 4 – j3, Z3 = –2 + j3 and Z4 = –5 – j

(a) 1

2

ZZ

(b) 1 3

2 4

Z ZZ Z

+−

(a) 21

2 22

1 2 (1 2)(4 3) 4 3 8 6 4 3 8 6 2 114 3 (4 3)(4 3) 4 3 25 25

Z j j j j j j j j jZ j j j

+ + + + + + + + − − += = = = =

− − + + = 2 11

25 25j−

+

(b) 21 3

2 22 4

(1 2) ( 2 3) 1 5 ( 1 5)(9 2) 9 2 45 10 19 43(4 3) ( 5 ) 9 2 (9 2)(9 2) 9 2 85

Z Z j j j j j j j j jZ Z j j j j j

+ + + − + − + − + + − − + + − += = = = =

− − − − − − − + +

= 19 4385 85

j−+

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8. Evaluate in a + jb form, given Z1 = 1 + j2, Z2 = 4 – j3, Z3 = –2 + j3 and Z4 = –5 – j

(a) 1 3

1 3

Z ZZ Z+

(b) Z2 + 1

4

ZZ

+ Z3

(a) 1 3

1 3

Z ZZ Z+

= 2(1 2)( 2 3) 2 3 4 6 8

(1 2) ( 2 3) 1 5 1 5j j j j j j

j j j j+ − + − + − + − −

= =+ + − + − + − +

= 2

2 2

( 8 )( 1 5) 8 40 5 3 41( 1 5)( 1 5) 1 5 26

j j j j j jj j

− − − − + + + += =

− + − − + = 3 41

26 26j+

(b) 12 3

4

ZZ ZZ

+ + = (4 – j3) + 1 25

jj

+− −

+ (–2 + j3) = 4 – j3 +2 2

(1 2)( 5 )5 1j j+ − ++

– 2 + j3

= 4 – j3 + 25 10 2

26j j j− + − + – 2 + j3

= 4 – j3 + 7 926

j− − – 2 + j3 = 2 – 7 926 26

j−

= 52 7 926 26 26

j− − = 45 926 26

j−

9. Evaluate (a) 11

jj

−+

(b) 11 j+

(a) 2

2 2

1 (1 )(1 ) 1 21 (1 )(1 ) 1 1 2

j j j j j j jj j j

− − − − − + −= = =

+ + − + = 0 – j1 = – j

(b) 2 2

1 (1)(1 ) 1 11 (1 )(1 ) 1 1 2

j j jj j j

− − −= = =

+ + − + = 1 1

2 2j−

10. Show that 25 1 2 2 52 3 4

j jj j

− + −− + −

= 57 + j24

2

2 2

1 2 (1 2)(3 4) 3 4 6 8 11 2 11 23 4 3 4 25 25 25 25

j j j j j j j jj

+ + − − + − += = = = +

+ +

2

2

2 5 (2 5)( ) 2 5 5 2 5 2( ) 1

j j j j j j jj j j j

− − − += = = = +

− − −

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L.H.S. = 25 1 2 2 5 25 11 2 25 11 2(5 2) 5 22 3 4 2 25 25 2 25 25

j j j j jj j

− + − − = − + − + = − − + − + −

= 25 11 125 2 50 25 114 482 25 25 2 25 25

j j − − − + = − − −

= 25 114 25 482 25 2 25

j − − +

= 57 + j24 = R.H.S.

© 2014, John Bird

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1. Solve: (2 + j)(3 – j2) = a + jb

(2 + j)(3 – j2) = a + jb Hence, 6 – j4 + j3 – 2j 2 = a + jb i.e. 8 – j1 = a + jb Thus, a = 8 and b = –1

2. Solve: 21

jj

+−

= j(x + jy)

2 ( )1

j j x jyj

+= +

− hence, (2 )(1 ) ( )

(1 )(1 )j j j x jyj j

+ += +

− +

i.e. 2

22 2

2 21 1j j j jx j y+ + +

= ++

i.e. 1 32

j jx y+= −

i.e. 1 32 2

j+ = –y + jx

Hence, x = 32

and y = 12

−

3. Solve: (2 – j3) = ( )a b+

(2 3) ( )j a jb− = +

Squaring both sides gives: ( )22 3j a jb− = +

(2 – j3)(2 – j3) = a + jb

i.e. 4 – j6 – j6 + 2j 9 = a + jb

i.e. –5 – j12 = a + jb

Hence, a = –5 and b = –12

4. Solve: (x – j2y) – (y – jx) = 2 + j

© 2014, John Bird

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(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 + 1j Cω

= 10 + j(4)(5) + 1(4)(0.04)j

= 10 + j20 + 6.25j

= 10 + j20 + 6.25( )( )

jj j

−−

= 10 + j20 – 2

6.25j−

= 10 + j20 – j6.25

= 10 + j13.75

© 2014, John Bird

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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 = 2 24 2+ = 4.472

Argument, θ = 1 4tan2

− = 63.43°

(b) –5 – j2 lies in the third quadrant as shown below

Modulus, r = 2 25 2+ = 5.385

α = 1 2tan5

− = 21.80°

Hence, argument, θ = –(180° – 21.80°) = –158.20°

(c) j(2 – j) = j2 – 2j = j2 + 1 or 1 + j2

1 + j2 lies in the first quadrant as shown below.

Modulus, r = 2 21 2+ = 2.236

Argument, θ = 1 2tan1

− = 63.43°

© 2014, John Bird

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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 = 2 22 3 13+ =

and 1 3tan 56.31 or 56 19 '2

θ − = = ° °

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, –4 = 4 180∠ ° in polar form (c) –6 + j From the diagram below, r = 2 26 1 37+ =

and 1 1tan 9.466

α − = = °

thus θ = 180° – 9.46° = 170.54°

Thus, –6 + j = 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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(a) –j3 From the diagram below, r = 3 and θ = –90°

Hence, – j3 = 3∠–90° in polar form

(b) ( )32 j− + = (–2 + j)(–2 + j)(–2 + j) = (4 – j2 – j2 + 2j )(–2 + j)

= (3 – j4)(–2 + j) = –6 + j3 + j8 – 2j 4 = –2 + j11

From the diagram below, r = 2 22 11 125+ = and 1 11tan 79.702

α − = = °

and θ = 180° – 79.70° = 100.30°

Hence, ( )32 j− + = –2 + j11 = 125 100.30∠ ° in polar form

(c) 3 (1 )j j− = (j)( 2j )(1 – j) = –j(1 – j) = –j + 2j = –1 – j

From the diagram below, r = 2 21 1 2+ = and 1 1tan 451

α − = = °

and θ = 180° – 45° = 135°

Hence, 3 (1 )j 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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(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 27 152 15 2

∠ °= ∠ °− − °

∠− ° = 3.2∠42°

(b) 5 30 4 80 10 40∠ °× ∠ °÷ ∠− ° = 5 30 4 80 5 4 (30 80 40 )10 40 10∠ °× ∠ ° ×

= ∠ °+ °− − °∠− °

= 2∠150°

8. Evaluate in polar form: (a) 46π

∠ + 38π

∠ (b) 2∠120° + 5.2∠58° – 1.6∠– 40°

(a) 4 36 8π π

∠ + ∠ = 4cos 4sin 3cos 3sin6 6 8 8

j jπ π π π + + +

= (3.464 + j2) + (2.772 + j1.148)

= 6.236 + j3.148

From the diagram below, r = 2 26.236 3.148 6.986+ =

© 2014, John Bird

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and 1 3.148tan 26.79 or 0.467 rad6.236

θ − = = °

Hence, 4 3

6 8π π

∠ + ∠ = 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 = 2 20.530 7.170 7.190+ =

and 1 7.170tan 85.770.530

θ − = = °

Hence, 2 120 5.2 58 1.6 40∠ °+ ∠ °− ∠− ° = 7.190∠85.77°

© 2014, John Bird

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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, LX = 8 Ω (since the j

term is positive)

LX = 2πfL = 8 hence, inductance, L = 8 82 2 (50)fπ π

= = 0.0255 H or 25.5 mH

(b) If Z = (2 – j3) Ω then resistance, R = 2 Ω and capacitive reactance, CX = 3 Ω (since the j

term is negative)

12

CXfCπ

= = 3 hence, capacitance, C = 3 61 1 1.061 10 or 1061 102 (3) 2 (50)(3)fπ π

− −= = × ×

= 1061 µF

(c) If Z = j14 Ω i.e. Z = (0 + j14) Ω then resistance, R = 0 Ω and LX = 14 Ω

i.e. 2πfL = 14 hence, inductance, L = 142 (50)π

= 0.04456 H or 44.56 mH

(d) If Z = 8 60∠− °Ω = 8 cos(–60°) + j8 sin(–60°) = (4 – j6.928) Ω

Hence, resistance, R = 4 Ω and CX = 6.928 Ω

i.e. 1 6.9282 fCπ

= and capacitance, C = 61 459.4 102 (50)(6.928)π

−= × = 459.4 µF

2. Two impedances, Z1 = (3 + j6) Ω and Z2 = (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, 1 2TOTALZ Z Z= + = (3 + j6) + (4 – j3) = (7 + j3) Ω

= 2 2 1 37 3 tan7

− + ∠

= 7.616∠23.20° Ω

Since voltage V = 120∠0° V, then current, I = 120 07.616 23.20

VZ

∠ °=

∠ ° = 15.76∠– 23.20° A

i.e. the current is 15.76 A and is lagging the voltage by 23.20°

© 2014, John Bird

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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 TZ is given by:

2 2 2 21 2

1 1 1 1 1 3 6 4 3 3 6 4 33 6 4 3 3 6 4 3 45 45 25 25T

j j j jZ Z Z j j

− += + = + = + = − + +

+ − + +

i.e. 1 admittance, TT

YZ

= = 0.22667 – j0.01333 = 0.2271∠–3.37° siemen

Current, I = (120 0 )(0.2271 3.37 ) 27.25 3.37TT

V VYZ

= = ∠ ° ∠− ° = ∠− ° A

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, LX = 2πfL = 2π(50)(0.10) = 31.416 Ω

and capacitive reactance, ( )( )6

1 12 2 50 160 10

CXf Cπ π −

= =×

= 19.894 Ω

Hence, impedance, Z = R + j( L CX X− ) = 12 + j(31.416 – 19.894) = (12 + j11.52) Ω = 16.64∠43.83° Ω

Current flowing, I = 240 016.64 43.83

VZ

∠ °=

∠ ° = 14.42∠– 43.83° A

Phase angle = 43.83° lagging (i.e. I lags V by 43.83°)

Power factor = cos ϕ = cos 43.83° = 0.721

© 2014, John Bird

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5. For the circuit shown, determine the current I flowing and its phase relative to the applied voltage.

2 2 2 21 2 3

1 1 1 1 1 1 1 30 20 40 50 130 20 40 50 25 30 20 40 50 25T

j jZ Z Z Z j j

+ −= + + = + + = + +

− + + +

= 30 20 40 50 11300 1300 4100 4100 25

j j+ + − +

i.e. 1 admittance, TT

YZ

= = 0.07283 + j0.00319 = 0.0729∠2.51° S

Current, I = (200 0 )(0.0729 2.51 ) 14.6 2.51TT

V VYZ

= = ∠ ° ∠ ° = ∠ ° A

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 = 5 0 9 135 12 240A B CF F F+ + = ∠ °+ ∠ °+ ∠ °

= (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)

© 2014, John Bird

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7. A delta-connected impedance ZA is given by:

ZA = 1 2 2 3 3 1

2

Z Z Z Z Z ZZ

+ +

Determine ZA in both Cartesian and polar form given Z1 = (10 + j0) Ω, Z2 = (0 – j10) Ω and

Z3 = (10 + j10) Ω

1 2 2 3 3 1

2

(10 0)(0 10) (0 10)(10 10) (10 10)(10 0)(0 10)

AZ Z Z Z Z Z j j j j j jZ

Z j+ + + − + − + + + +

= =−

= 2100 100 100 100 100 200 100 200 10010 10 10 10

j j j j j jj j j j

− − − + + −= = −

− − − −

= 200 10 (10 20)10

j j+ = + Ω

From the diagram below, r = 2 210 20+ = 22.36 and 1 20tan 63.4310

θ − = = °

Hence, (10 20) 22.36 63.43AZ j= + Ω = ∠ °Ω 8. In the hydrogen atom, the angular momentum p of the de Broglie wave is given by

pψ = –2jhπ

(± jmψ). Determine an expression for p.

If ( )2j hp j mπ

Ψ = − ± Ψ

then p = ( ) ( )( )2

2 2 2j h j m j h hj m j mπ π π

± Ψ − = − ± = − ± Ψ

= ( )2h mπ

± = 2m hπ

±

© 2014, John Bird

775

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 = P Qv v− = (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 = Q Pv v− = (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)

From the diagram below, r = 1.488 and 1 1.464tan 79.630.268

α − = = °

and 180 79.63 100.37θ = °− ° = °

© 2014, John Bird

776

Hence, P – Q – R = 1.488∠100.37°

11. In a Schering bridge circuit, ( )Xx X CZ R jX= − , 22 CZ jX= − , ( )( )( )

3

3

33

3

C

C

R jXZ

R jX−

=−

and

4 4Z R= where 12

CXfCπ

= . At balance: ( )( ) ( )( )3 2 4XZ Z Z Z= .

Show that at balance 3 4

2X

C RRC

= and 2 3

4X

C RCR

=

Since ( )( ) ( )( )3 2 4XZ Z Z Z=

then ( )( ) ( )( )3

2

3

34

3( )X

CX C C

C

R jXR jX jX R

R jX −

− = − −

Thus, ( )( )( )( )( )

3 2

3

3 4

3( )X

C CX C

C

R jX jX RR jX

R jX− −

− =−

i.e. ( )( ) ( )( )

2 3 2

3 3

23 4 4

3 3( )X

C C CX C

C C

j R X R j X X RR jXR jX R jX−

− = +− −

i.e. ( )

2 2

3

4 4

3( )

( )X

C CX C

C

X R X RR jXX R j

− = −−

= 2 2

3

4 4

3

C C

C

X R X RX j R

+

i.e. ( )XX CR jX− = 2 2

3

4 4

3

C C

C

X R X RjX R

−

Equating the real parts gives: 2

3

44 32

42

3

122

1 22

CX

C

RX R fCfCR RX fC

fC

πππ

π

= = =

i.e. 3 4

2X

C RRC

=

Equating the imaginary parts gives: 2 4

3X

CC

X RXR

− = −

i.e. 4

42

3 2 3

11 2

2 2X

RRfC

fC R fC Rπ

π π= =

from which, 2 3

4X

C RCR

=

© 2014, John Bird

777

12. An amplifier has a transfer function T given by T = ( )4

5001 5 10jω −+ ×

where ω is the angular

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).

When ω = 2000 rad/s , transfer function T = ( )4 4

500 500 5001 5 10 1 (2000)(5 10 ) 1 1j j jω − −

= =+ × + × +

Hence, T = 2 2

500 (500)(1 1) 500 500 500 500 500 5001 1 (1 1)(1 1) 1 1 2 2 2

j j j jj j j

− − −= = = = −

+ + − +

= 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 0

tanhSS

VI PLZ

= . Calculate the

value of the sending current, in polar form, given 200 VSV = , 0 560 420Z j= + Ω , P = 0.20

and L = 10

Sending current, ( )0

200 200 tanh 2tanh tanh 0.20 10(560 420) (560 420)

SS

VI PLZ j j

= = × =+ +

= 2 2

192.8 (192.8)(560 420) (192.8)(560 420)(560 420) (560 420)(560 420) 560 420

j jj j j

− −= =

+ + − +

= ( )(192.8) 700 36.870.275 36.87

490000∠− °

= ∠− ° A

i.e. the sending end current, SI = 275 36.87∠− ° mA

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