Given that a circuit element has a complex number phasor voltage Us" = 30 m'v' < FD" across it w the circuit element has a complex number impedance Z = 2 "'59 a: 130" Find the following: a) Find the complex number phasor current I" b} Write the time domain sinusoidal math expressions for voltage v5 and current i as functions of time t {The angular velocity can be expressed as m in 1vour time domain sinusoidal math expressions for part bl c) State the phase relationship between the voltage v5 and currenti sinusoidal math expressions of part b

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The given data ! 9) Complex number Voltage phasor ( V. )= 30 mV/ 70 Complex number impedence phasor ( 2 )= 2kn /130 ( 9) Complex number Current phasor ( I ) = ? we know G = VS 1 = 30 X10 270 2 X103 / 180 1 = 30X10 2 X10 3 (270- 130' ) I = 15 x 10 6 / - 60 I = 15 4AL60 ( 07 ) 1 = 15 /60 MA ( b) Time domain Sinusoidal math expression for Voltage Us given that V = 30 mV / 70 equation in the form of Vs = VOL. V = VmLov -2 - time domain Sinusoidal math expression for voltage is V (t ) = Vm sin (wot + 8) v ) -- from ean D and 2 Vm = 30 mv Qv = 70 Now substitute above values in ean (g : 80x 10 = 30 mV Vs ( t ) = 30*10 sin (wt +70 ) vistime domain equation / expression for VS = 30 mv / 70 Similarly for current : we got Is = 15 414 - 60 -> 1 = Im LP, I'm = 15 MA Di = - 60 -> Time domain Sinusoidal math expression for Current i (t ) = Im sin ( wt + $; ) - Now Substitute Im, b; values in can i(t ) = 15 * 10 6 sin ( wot + ( - 68 ) ) ( # 154A = 15 * 10 6A . . i(t ) = 15 x10 sin ( wt - 60 ' ) A from part (6) it is clear that the voltage having voltage phasor Angle DV = 70 and current phase di= - 60 - ) to know more about phase relationship between voltage and Current lam giving graphical representation given below.V ( t ) = 30 *10 - sin ( wt + 70' ) , PA)= 15x106 sin ( wt- 60" ) 7OK -60 TT 1/2 2 IT 60 from above graph : The voltage Leads the Current by 130 degrees - The Current lagging the voltage by 130 degrees 10 The given dat : I = 8. 9 MA / 120 2 = 4. 7 km / 20 ( a ) Complex Number phasor voltage v = 1.2 V = 8 . 9 X 10 3 / 12 0 X 4. 7 X 103 / 20' = 8.9x183 x 4.7*103 /12+20 V = 41. 83 / 32" V ( b) Time domain sinusoidal math expression for voltage !. V ( t ) = Vm sin (wt+ Qv ) we have V = 41.83 / 32 = > V = Vm LDv Vm = 41.83 , Dy = 32"Substitute above values in egno . V( t ) = 41.83 sin ( wt + 320 ) Similarly, -) Time domain sinusoidal
math expression for Current : it ) = I'm sin (wot thi) - 0 We have I = 8.9 mA / 120 = I = Im LDi I'm = 8.
9*10' , di = 12 Now, Substitute above values in each 1 goo i (t ) = 8. 9 *10 sin (wot + 129 ) A