CHAPTER 45 COMPLEX NUMBERS

Size: px

Start display at page:

Download "CHAPTER 45 COMPLEX NUMBERS"

Emory Paul
6 years ago
Views:

1 CHAPTER 45 COMPLEX NUMBERS EXERCISE 87 Page 50. Solve the quadratic equation: x Since x then x 5 x 5 ( )(5) 5 j 5 from which, x ± j5. Solve the quadratic equation: x x + 0 Since x x + 0 then [ ] ± ( ) 4()() ± 4 ± ( )(4) ± ( ) (4) ± j (4) x () ± j 4 ± j ± j ± j 3. Solve the quadratic equation: x 4x Since x 4x then [ ] 4 ± ( 4) 4()(5) 4± 4 4 ± ( )(4) 4 ± ( ) (4) 4 ± j (4) x () 4 ± j 4 4 ± j ± j ± j 4. Solve the quadratic equation: x 6x Since x 6x then [ ] 6 ± ( 6) 4()(0) 6± 4 6 ± ( )(4) 6 ± ( ) (4) 6 ± j (4) x () , John Bird

2 6 ± j 4 6 ± j 3 ± j 5. Solve the quadratic equation: x x + 0 Since x x + 0 then [ ] ± ( ) 4()() ± 4 ± ( )(4) ± ( ) (4) ± j (4) x () ± j 4 ± j 0.5 ± j Solve the quadratic equation: x 4x Since x 4x then [ ] 4 ± ( 4) 4()(8) 4 ± 6 4 ± ( )(6) 4 ± ( ) (6) 4 ± j (6) x () 4 ± j 6 4 ± j 4 ± j 7. Solve the quadratic equation: 5x 0x + 0 Since 5x 0x + 0 then [ ] 0 ± ( 0) 4(5)() 0 ± 00 0 ± ( )(00) 0 ± ( ) (00) x (5) ± j (00) ± j ± j 0. ± j Solve the quadratic equation: x + 3x Since x + 3x+ 4 0 then [ ] 3 ± 3 4()(4) 3 ± 3 3 ± ( )(3) 3 ± ( ) (3) 3 ± j (3) x () , John Bird

3 3 ± j 3 or ( ± j.99) Solve the quadratic equation: 4t 5t Since 4t 5t then [ ] 5 ± ( 5) 4(4)(7) 5 ± 87 5 ± ( )(87) 5 ± ( ) (87) 5 ± j (87) t (4) ± j 87 or (0.65 ± j.66) Evaluate (a) j 8 (b) j 7 (c) 4 j 3 (a) 8 j ( j ) 4 ( ) (b) 7 6 ( ) ( ) j j j j j j j Hence, j j j j j7 j j j( j) j ( ) j (c) ( ) 6 j3 j j j j j ( ) 6 j Hence, 4 j 3 ( j) j j j j j( j) j , John Bird

4 EXERCISE 88 Page 53. Evaluate (a) (3 + j) + (5 j) and (b) ( + j6) (3 j) and show the results on an Argand diagram. (a) (3 + j) + (5 j) (3 + 5) + j( ) 8 + j (b) ( + j6) (3 j) + j6 3 + j ( 3) + j(6 + ) 5 + j8 (8 + j) and ( 5 + j8) are shown on the Argand diagram below.. Write down the complex conjugates of (a) 3 + j4, (b) j. (a) The complex conjugate of 3 + j4 is: 3 j4 (b) The complex conjugate of j is: + j 3. If z + j and w 3 j evaluate (a) z + w (b) w z (c) 3z w (d) 5z + w (e) j(w 3z) (f) jw jz (a) z + w ( + j) + (3 j) + j + 3 j 5 (b) w z (3 j) ( + j) 3 j j j (c) 3z w 3( + j) (3 j) 6 + j3 6 + j j5 (d) 5z + w 5( + j) + (3 j) 0 + 5j + 6 j 6 + j3 (e) j(w 3z) j[(6 j) (6 +j3)] j[6 j 6 j3] j( j5) j 5 ( )5 5 (f) jw jz j(3 j) j( + j) j6 j j j j6 ( ) j ( ) j6 + j j4 4. Evaluate in a + jb form, given + j, 4 j3, 3 + j3 and 4 5 j (a) + 3 (b) , John Bird

5 (a) j + 4 j3 ( + j3) + j + 4 j3 + j3 ( ) + j( 3 3) 7 j4 (b) + 4 (4 j3) ( + j) + ( 5 j) 4 j3 j 5 j (4 5) + j( 3 ) j6 5. Evaluate in a + jb form, given + j, 4 j3, 3 + j3 and 4 5 j (a) (b) 3 4 (a) ( + j)(4 j3) 4 j3 + j8 j 6 4 j3 + j j5 (b) 3 4 ( + j3)( 5 j) 0 + j j5 j j j j3 6. Evaluate in a + jb form, given + j, 4 j3, 3 + j3 and 4 5 j (a) (b) 3 (a) ( + j)( + j3) + ( 5 j) + j3 j4 + j 6 5 j + j3 j4 6 5 j 3 j (b) 3 ( + j)(4 j3)( + j3) (4 j3 + j8 j 6)( + j3) ( 0 + j5)( + j3) 0 + j30 j0 + j j30 j j0 7. Evaluate in a + jb form, given + j, 4 j3, 3 + j3 and 4 5 j (a) (b) (a) j j j j j j j j j j j j + (+ )(4+ 3) (4 3)(4 + 3) j 5 5 (b) j + + j + j + j + j j + j + j + j j j j j + j + 4 ( ) ( 3) 5 ( 5)(9 ) (4 3) ( 5 ) 9 (9 )(9 ) j , John Bird

6 8. Evaluate in a + jb form, given + j, 4 j3, 3 + j3 and 4 5 j (a) (b) (a) (+ j)( + j3) + j3 j4+ j6 8 j (+ j) + ( + j3) + j5 + j5 ( 8 j)( j5) 8 + j40 + j+ j5 3 + j4 ( + j5)( j5) j (b) + + (4 j3) + + j 5 j 3 4 (+ j)( 5 + j) + ( + j3) 4 j j3 4 j j j0 j + j3 4 j3 + 7 j9 6 + j3 7 j j 9 45 j Evaluate (a) + j j (b) + j (a) j ( j)( j) j j+ j j j ( j)( j) j j (b) ()( j) j j j ( j)( j) j 0. Show that 5 + j j5 3+ j4 j 57 + j4 + j ( + j)(3 j4) 3 j4 + j6 j8 + j + j 3 + j j5 ( j5)( j) j j5 5+ j j j( j) j 5+ j 76 04, John Bird

7 L.H.S. 5 + j j5 5 5 j (5 j) j 3+ j4 j j j j 57 + j4 R.H.S , John Bird

8 EXERCISE 89 Page 54. Solve: ( + j)(3 j) a + jb ( + j)(3 j) a + jb Hence, 6 j4 + j3 j a + jb 8 j a + jb Thus, a 8 and b. Solve: + j j j(x + jy) + j j ( x + jy) j hence, ( + j)( + j) j( x + jy) ( j)( + j) Hence, x 3 + j+ j+ j + jx + + j3 jx y j y 3 + j y + jx and y 3. Solve: ( j3) ( a+ b) Squaring both sides gives: ( ) ( j3) ( a + jb) j3 a + jb ( j3)( j3) a + jb 4 j6 j6 + j 9 a + jb 5 j a + jb Hence, a 5 and b 4. Solve: (x jy) (y jx) + j , John Bird

9 (x jy) (y jx) + j Hence, (x y) + j( y + x) + j x y () and x y () () () gives: y Substituting in () gives: x from which, x 3 5. If R + jωl + /jωc, express in (a + jb) form when R 0, L 5, C 0.04 and ω 4 R + jωl + jω C 0 + j(4)(5) + j (4)(0.04) 0 + j j 0 + j ( j ) j( j) 0 + j j0 j6.5 j 0 + j , John Bird

j4 lies in the first quadrant as shown below Modulus, r 4 + 4.

43 (b) 5 j lies in the third quadrant as shown below Modulus, r 5 + 5.

10 EXERCISE 90 Page 57. Determine the modulus and argument of (a) + j4 (b) 5 j (c) j( j) (a) + j4 lies in the first quadrant as shown below Modulus, r Argument, θ tan (b) 5 j lies in the third quadrant as shown below Modulus, r α tan.80 5 Hence, argument, θ (80.80 ) 58.0 (c) j( j) j j j + or + j + j lies in the first quadrant as shown below. Modulus, r +.36 Argument, θ tan , John Bird

11 . Express in polar form, leaving answers in surd form: (a) + j3 (b) 4 (c) 6 + j (a) + j3 From the diagram below, r and θ 3 tan 56.3 or 56 9' Hence, + j in polar form (b) j0 and is shown in the diagram below, where r 4 and θ 80 Hence, in polar form (c) 6 + j From the diagram below, r and α tan thus θ Thus, 6 + j Express in polar form, leaving answers in surd form: (a) j3 (b) ( + j) 3 (c) j 3 ( j) , John Bird

12 (a) j3 From the diagram below, r 3 and θ 90 Hence, j in polar form (b) ( + j) 3 ( + j)( + j)( + j) (4 j j + j )( + j) (3 j4)( + j) 6 + j3 + j8 j 4 + j From the diagram below, r + 5 and α tan and θ Hence, ( ) 3 j + + j in polar form (c) j3( j) (j)( j )( j) j( j) j + j j From the diagram below, r + and α tan 45 and θ Hence, j 3 ( j) j Convert into (a + jb) form giving answers correct to 4 significant figures: (a) 5 30 (b) 3 60 (c) , John Bird

13 (a) cos 30 + j 5 sin j.500 (b) cos 60 + j 3 sin j.598 (c) cos 45 + j 7 sin j Convert into (a + jb) form giving answers correct to 4 significant figures: (a) 6 5 (b) 4 π (c) (a) cos 5 + j 6 sin j4.95 (b) 4 π 4 cos π + j sin π j0 (Note that π is radians) (c) cos( 0 ) + j 3.5 sin( 0 ).750 j Evaluate in polar form: (a) (b) (a) ( ) (b) (65 + ) Evaluate in polar form: (a) (b) (a) (b) ( ) π π 8. Evaluate in polar form: (a) (b) π π (a) cos π j4sin π 3cos π j3sin π ( j) + (.77 + j.48) j3.48 From the diagram below, r , John Bird

14 and θ 3.48 tan 6.79 or rad 6.36 π π Hence, or rad 6 8 (b) ( cos 0 + j sin 0 ) + (5. cos 58 + j5. sin 58 ) (.6 cos( 40 ) + j.6 sin( 40 )) ( + j.73) + ( j4.40) (.6 j.08) + j j j j7.70 From the diagram below, r and θ 7.70 tan Hence, , John Bird

15 EXERCISE 9 Page 59. 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) ( j3) Ω (c) j4 Ω (d) 8 60 Ω (a) If (3 + j8) Ω then resistance, R 3 Ω and inductive reactance, X L 8 Ω (since the j X L πfl 8 hence, inductance, L H or 5.5 mh π f π(50) term is positive) (b) If ( j3) Ω then resistance, R Ω and capacitive reactance, X C 3 Ω (since the j X C term is negative) π fc 3 hence, capacitance, C or π f (3) π(50)(3) 06 µf (c) If j4 Ω (0 + j4) Ω then resistance, R 0 Ω and X L 4 Ω πfl 4 hence, inductance, L 4 π (50) H or mh (d) If 8 60 Ω 8 cos( 60 ) + j8 sin( 60 ) (4 j6.98) Ω Hence, resistance, R 4 Ω and X C 6.98 Ω π fc 6.98 and capacitance, C µf π (50)(6.98). Two impedances, (3 + j6) Ω and (4 j3) Ω are connected in series to a supply voltage of 0 V. Determine the magnitude of the current and its phase angle relative to the voltage In a series circuit, total impedance, TOTAL + (3 + j6) + (4 j3) (7 + j3) Ω Since voltage V 0 0 V, then current, I V tan Ω the current is 5.76 A and is lagging the voltage by A , John Bird

16 3. If the two impedances in Problem are connected in parallel, determine the current flowing and its phase relative to the 0 V supply voltage. In a parallel circuit shown below, the total impedance T is given by: 3 j6 4+ j j + + j j j T T admittance, YT j siemen V Current, I VYT (0 0 )( ) A T the current is 7.5 A and is lagging the voltage by A series circuit consists of a Ω resistor, a coil of inductance 0.0 H and a capacitance of 60 µf. Calculate the current flowing and its phase relative to the supply voltage of 40 V, 50 Hz. Determine also the power factor of the circuit. R Ω, inductive reactance, X L πfl π(50)(0.0) 3.46 Ω and capacitive reactance, X C π fc π Ω 6 ( )( ) Hence, impedance, R + j( XL XC) + j( ) ( + j.5) Ω Ω Current flowing, I V A Phase angle lagging ( I lags V by ) Power factor cos ϕ cos , John Bird

17 5. For the circuit shown, determine the current I flowing and its phase relative to the applied voltage j0 40 j j j T T j + j admittance, Y T j S V Current, I VYT (00 0 )( ) A T the current is 4.6 A and is leading the voltage by.5 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 35 to force A, Force C, N acting at an angle of 40 to Force A. Resultant force F + F + F A B C (5 + j0) + ( j6.364) + ( 6 j0.39) 5 + j j j0.39) j or N Hence, the magnitude of the force that has the same effect as the three forces acting separately is: 8.39 N and its direction is to the horizontal ( from Force A) , John Bird

18 7. A delta-connected impedance A is given by: A Determine A in both Cartesian and polar form given (0 + j0) Ω, (0 j0) Ω and 3 (0 + j0) Ω A (0 + j0)(0 j0) + (0 j0)(0 + j0) + (0 + j0)(0 + j0) (0 j0) j00 j00 j j00 00 j00 00 j00 j0 j0 j0 j0 j (0 + j 0) Ω 0 From the diagram below, r and θ 0 tan Hence, (0 + j0) Ω 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. π jh π If pψ ( ± jmψ) jh ± jmψ jh h π Ψ π π then p ( ± jm ) ( j )( ± m) h π ± mh ± π ( m) , John Bird

19 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 (00 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) (00 j600) j j j the velocity of P relative to Q is 9 km/h at (b) The velocity of Q relative to P v Q v P (00 j600) (400 + j300) 00 j j j the velocity of Q relative to P is 9 km/h at Three vectors are represented by P, 30, Q, 3 90 and R, Determine in polar form the vectors represented by (a) P + Q + R, (b) P Q R. (a) P + Q + R (.73 + j) + (0 + j3) + ( j3.464) ( j0.536) (b) P Q R (.73 + j) (0 + j3) ( j3.464) ( j.464) From the diagram below, r.488 and α.464 tan and θ , John Bird

20 Hence, P Q R In a Schering bridge circuit, x ( RX jx C X ) R where 4 4 X C π fc, jx C,. At balance: ( )( ) ( )( ) X ( R3 )( jx C3 ) ( R3 jx C3 ) and Show that at balance R X C3R4 and C C X C R R 3 4 Since ( X )( 3) ( )( 4) then Thus, ( R3 )( jx C3 ) ( RX jx CX ) jx C R R3 jx C3 ( R jx ) X CX ( )( 4 ) ( R3 jx C3)( jx C)( R4) ( R3 )( jx C3 ) jr X R j X 3X R ( R jx ) + X C X 3 C 4 C C 4 ( R3)( jx C3) ( R3)( jx C3) XCR4 XCR4 ( RX jx CX ) X R ( j) C 3 C 3 ( 3 ) XCR X R + X jr 4 C 4 3 ( R jx ) X C X XCR X R j X R C 3 4 C 4 3 Equating the real parts gives: R X R X C R π fc fc X C3 π fc π fc 4 4 π 3 3 R 4 CR 3 4 RX C XC R4 Equating the imaginary parts gives: X CX R3 R4 π fc R4 π fc R π fc R X 3 3 from which, C X CR R , John Bird

21 . An amplifier has a transfer function T given by T jω ( ) 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 000 rad/s, determine the gain and the phase (in degrees). When ω 000 rad/s, transfer function T ( ) + jω j(000)(5 0 4) + j Hence, T 500 (500)( j) 500 j j j + j (+ j)( j) + 50 j Hence, the gain of the amplifier and the phase is The sending end current of a transmission line is given by I S VS tanh PL. Calculate the value of the sending current, in polar form, given V 00 V, j40 Ω, P 0.0 and L 0 S 0 V tanh (560 + j40) (560 + j40) S Sending current, IS tanh PL tanh ( 0.0 0) (9.8)(560 j40) (9.8)(560 j40) (560 + j40) (560 + j40)(560 j40) ( ) (9.8) A the sending end current, I S ma , John Bird

Single Phase Parallel AC Circuits

Single Phase Parallel AC Circuits 1 Single Phase Parallel A.C. Circuits (Much of this material has come from Electrical & Electronic Principles & Technology by John Bird) n parallel a.c. circuits similar