APPENDIX A Exercises EA.1 Given Z = − j and Z = + j 6, we have: Z + Z = 10 + j Z − Z = −6 − j Z Z = 16 − j 24 + j 12 − j 18 = 34 − j 12 − j − j 16 − j 12 − j 24 + j 18 Z1 / Z2 = × = = −0.02 − j 0.36 8+ j6 8− j6 100 EA.2 Z = 15∠45 o = 15 cos( 45 o ) + j 15 sin( 45 o ) = 10.6 + j 10.6 Z = 10∠ − 150 o = 10 cos( −150 o ) + j 10 sin( −150 o ) = −8.66 − j Z = 5∠90 o = cos(90 o ) + j sin(90 o ) = j EA.3 Notice that Z1 lies in the first quadrant of the complex plane Z = + j = 32 + ∠ arctan( / 3) = 5∠53.13o Notice that Z2 lies on the negative imaginary axis Z = − j 10 = 10∠ − 90 o Notice that Z3 lies in the third quadrant of the complex plane Z = −5 − j = 52 + 52 ∠(180 o + arctan( −5 / − 5)) = 7.07 ∠225 o = 7.07 ∠ − 135 o EA.4 Notice that Z1 lies in the first quadrant of the complex plane Z = 10 + j 10 = 10 + 10 ∠ arctan(10 / 10) = 14.14∠45 o = 14.14 exp( j 45 o ) Notice that Z2 lies in the second quadrant of the complex plane Z = −10 + j 10 = 10 + 10 ∠(180 o + arctan( −10 / 10)) = 14.14∠135 o = 14.14 exp( j 135 o ) EA.5 Z 1Z = (10∠30 o )(20∠135 o ) = (10 × 20)∠(30 o + 135 o ) = 200∠(165 o ) Z / Z = (10∠30 o ) /(20∠135 o ) = (10 / 20)∠(30 o − 135 o ) = 0.5∠( −105 o ) Z − Z = (10∠30 o ) − (20∠135 o ) = (8.66 + j 5) − ( −14.14 + j 14.14) = 22.8 − j 9.14 = 24.6∠ − 21.8o Z + Z = (10∠30 o ) + (20∠135 o ) = (8.66 + j 5) + ( −14.14 + j 14.14) = −5.48 + j 19.14 = 19.9∠106o Problems PA.1 Given Z = + j and Z = − j 3, we have: Z1 + Z2 = + j Z − Z = −2 + j Z Z = − j + j 12 − j = 17 + j Z1 / Z2 = PA.2 + j + j − + j 18 = = 0.04 + j 0.72 × − j3 + j3 25 Given that Z = − j and Z = + j 3, we have: Z1 + Z2 = + j Z − Z = −1 − j Z1 Z2 = + j − j − j = − j Z1 / Z2 = − j2 − j3 − − j = = −0.3077 − j 0.5385 × + j3 − j3 13 PA.3 Given that Z = 10 + j and Z = 20 − j 20, we have: Z + Z = 30 − j 15 Z − Z = −10 + j 25 Z Z = 200 − j 200 + j 100 − j 100 = 300 − j 100 Z1 / Z2 = PA.4 PA.5 PA.6 10 + j 20 + j 20 100 + j 300 = = 0.125 + j 0.375 × 20 − j 20 20 + j 20 800 (a) Z a = − j = 7.071∠ − 45o = 7.071 exp(− j 45o ) (b) Z b = −10 + j = 11.18∠153.43o = 11.18 exp(j 153.43o ) (c) Z c = −3 − j = 5∠ − 126.87 o = exp(− j 126.87 o ) (d) Z d = − j 12 = 12∠ − 90 o = 12 exp(− j 90 o ) (a) Z a = 5∠45o = exp(j 45o ) = 3.536 + j 3.536 (b) Z b = 10∠120 o = 10 exp(j 120 o ) = −5 + j 8.660 (c) Z c = 15∠ − 90 o = 15 exp(− j 90 o ) = − j 15 (d) Z d = −10∠60 o = 10 exp(− j 120 o ) = −5 − j 8.660 (a) Z a = 5e j 30 = 5∠30 o = 4.330 + j 2.5 (b) Z b = 10e − j 45 = 10∠ − 45o = 7.071 − j 7.071 (c) Z c = 100e j 135 = 100∠135o = −70.71 + j 70.71 o o o PA.7 (d) Z d = 6e j 90 = 6∠90 o = j (a) Z a = + j + 10∠30 o = 13.66 + j 10 (b) Z b = 5∠45o − j 10 = 3.536 − j 6.464 (c) Zc = 10∠45 o 10∠45 o 2∠ − 8.13o = 1.980 − j 0.283 = + j4 5∠53.13o (d) Zd = 15 = 3∠ − 90 o = − j o 5∠90 o ... Problems PA.1 Given Z = + j and Z = − j 3, we have: Z1 + Z2 = + j Z − Z = −2 + j Z Z = − j + j 12 − j = 17 + j Z1 / Z2 = PA.2 + j + j − + j 18 = = 0.04 + j 0.72 × − j3 + j3 25 Given that Z = − j and... j 3, we have: Z1 + Z2 = + j Z − Z = −1 − j Z1 Z2 = + j − j − j = − j Z1 / Z2 = − j2 − j3 − − j = = −0.3077 − j 0.5385 × + j3 − j3 13 PA.3 Given that Z = 10 + j and Z = 20 − j 20, we have: Z +... − j 200 + j 100 − j 100 = 300 − j 100 Z1 / Z2 = PA.4 PA.5 PA.6 10 + j 20 + j 20 100 + j 300 = = 0.125 + j 0.375 × 20 − j 20 20 + j 20 800 (a) Z a = − j = 7.071∠ − 45o = 7.071 exp(− j 45o ) (b)