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1 SINUSOIDAL STEADY STATE CIRCUIT ANALYSIS 1. Introduction A sinusoidal current has the following form: where I m is the amplitude value; ω=2 πf is the angular frequency; φ is the phase shift. i (t )=I m.sin (ωt +φ) Each sine could be represented with a phaser and a complex number: i (t )=I m.sin ( ωt+φ ) I =I m.e jφ =I m cos (φ )+ j I m sin (φ ) Let s consider a series RLC circuit: The KVL for the circuit is: U IN =U R +U L +U C where the voltage drops on the resistor, inductor and capacitor are: U R =R.I U L = j X L I U C = j X C I X L and X C are the reactances of the inductor and the capacitor respectively: X L =ωl X C = 1 ωc Substituting the above equations in the KVL we obtain: U IN =U R +U L +U C =I (R+ j ( X L X C ))=I ( R+ jx ) where Z =R + jx is the complex impedance of the circuit. X =X L X C =ωl 1 ωc is the total reactance of the branche. Every complex number could be presented in rectangular or exponential/polar form, so the complex impedance can also be written as: where z is the magnitude of the impedance: Z =R + jx=z.e jφ =z φ 1 / 7

2 z= U IN = R 2 +X 2 I φ is the phase difference between the voltage and current in the branch: φ=atan X R 2. Problems Problem 1. Transform the sinusoids in complex form: i 1 (t )=10 sin (ωt +45 )[ A] i 2 (t )=12sin (ωt 90 ) [ A ] i 3 (t )=3sin (ωt +80 )[ A ] u 1 (t )=100 sin (ωt )[V ] u 2 (t )=50sin (ωt 35 )[V ] Solutions: I 1 =10 e j 45 [ A ] I 2 =12e j 90 [ A ] I 3 =3ej 80 [ A ] U 1 =100ej 0 =100 [V ] U 2 =50e j 35 [V ] Problem 2. Transform the phasors in sinusoidal form: I 1 =2.25 e j 30 [ A] I 2 =1.5e j 45 [ A ] U 1 =5+ j10[v ] U 2 = 10 j10[v ] Solutions: i 1 (t )=2.25sin (ωt +30 ) [ A ] i 2 (t )=1.5sin (ωt 45 ) [ A ] U =5+ 1 j10= e j arctg 10 5 =11,2e j 63 u 1 (t )=11,2 sin (ωt+63 ) [V ] U 10 = 10 j10= (10+ j10)= 2 (10) 2 +(10) 2 j arctg e 10 = 14,1e j 45 u 2 (t )=14,1 sin (ωt+45 ) [V ] 2 / 7

3 Problem 3. Obtain the reactances of the inductors and capacitors: L 1 =10mH at frequency f =50 Hz C 1 =0,5F at frequency f =110 Hz L 2 =150 μh at frequency f =10kHz L 3 =150 μh at frequency f =100kHz C 2 =10μF at frequency f =10 khz C 3 =10μF at frequency f =100 khz Solutions: X L1 =ω L 1 =2 πf L 1 =2π =3,14 [Ω ] X C 1 = 1 = 1 1 = =0,0029 [Ω ] ωc 1 2πf C 1 2 π.110.0,5 X L2 =ω L 2 =2 πf L 2 =2π =9,42 [Ω ] X L3 =ω L 3 =2πf L 3 =2π =94,2 [Ω ] X C 2 = 1 = 1 1 = =1,6 [Ω ] 6 ωc 2 2 πf C 2 2 π X C 3 = 1 = 1 1 = =0,16 [Ω ] 6 ωc 3 2 πf C 3 2π Problem 4. Obtain the current i (t ) if the frequency of the source is f =100 Hz. First we obtain the reactances of the reactive elements: X L =ωl=2 πfl=2. π.100.0,1=62 [Ω ] X C = 1 ωc = 1 2 πfc = 1 =159 [Ω ] 6 2. π Next we present the sinusoidal source as phasor: Е =10 e j20 [V ] Now we can draw an equivalent circuit with complex numbers: 3 / 7

4 We write the KVL equation for the loop: 10 e j =I (10+ j60 j159 ) 10ej I = 97,5 e =0,103 e j104 [ A] j 84 Finally we transform the phasor in sinusoidal form: I =0,103 e j 104 i (t )=0,103 sin ( ωt+104 ) [ A ] Problem 5. Obtain the complex currents in the circuit if the frequency of the source is f =20 Hz and obtain the active power, dissipated by the load R T. First we obtain the reactances: X L1 =ωl=2 πfl=2 π.20.0,5=63 [Ω ] X C 3 = 1 ωc = 1 2 πfc = 1 =8 [Ω ] 3 2π and present the sinusoidal source as phasor: i 1 (t )=1.sin (ωt +0) I 1 =1ej 0 =1 [ A ] Next we create an equivalent circuit with complex numbers: For the circuit we can write two equations using Kirchoff s laws method: 4 / 7

5 1=I 2 +I 3 0=10I + j 8 2 I 3 [ I 3] [ I 10 j 8] = [ 1 0] The determinants are: =[ =1. j8 10= 10+ j8 j 8] 2 =[ j 8] = j 8 3 =[ ] = 10 Then the complex currents are: I 2 = 2 = j j8 = 8 e j 90 12,8 e j 39 =0,63 e j 129 [ A] I = 3 3 = j8 = 10 12,8e j 39 =0,78ej 39 [ A ] The corresponding sinusoidal currents are: I 2 =0,63 e j 129 = i 2 (t )=0,63sin (ωt +129 )[ A ] I 3 =0,78ej 39 i 3 (t )=0,78sin (ωt +39 )[ A] Finally the dissipated power by the load R T is: P=( I 2 2 m 2 ) R T = ( 0,63 2 ) 2 10=1,98 [W ] Problem 6. Obtain the currents in the circuit and the powers of R 3 and L 3. We ll use nodal analysis. There are two nodes, so we let U 0 =0 and need 1 equation: I 1 =I 2 +I U 0 U j1 3 = U 1 U 0 7 j 20 + U 1 U j10 5 / 7

6 0,14+ j 0,14 0,14U 1 = j0,05u 1 +0,05U 1 j 0,05U 1 0,14+ j 0,14=0,19 U 1 U 1 =0,74+ j 0,74 Then our currents are: I 1 =0,14+ j0,14 0,14 U 1 =0,05.ej 45 [ A] I 2 = j 0,05U 1 = j 0,05(0,74+ j0,74 )=0,05. e j 135 [ A ] I =(0,05 j 3 0,05)U 1 =(0,05 j0,05)(0,74+ j 0,74)=0,07 [ A] The power dissipated in R 3 is active: P R3 =( I 2 3 m 2 ). R 3 = ( 0,07 2 ) 2.10=0,025 [W ] The power of the inductor is purely reactive: Q L 3 =( I 2 3 m 2 ). X L 3 = ( 0,07 2 ) 2.10=0,025 [VAr ] Problem 7. Obtain the complex currents in the circuit. The circuit has 2 unknown currents so we need 2 equations (using Kirchoff s laws): 1+I 2 =I 1 2,5+ j2,5=5i 2 +I 1 (2+ j 5 5 j ) In matrix form: [ ] = 1 [ 2,5+ j2,5] The determinants are: =[ ] = =7 2] [I 1 I 6 / 7

7 1 1 1 =[ 2,5+ j2,5 5 ] = j2.5=7,5+ j 2,5 2 =[ 1 1 =2.5+ j 2.5 2=0,5+ j 2,5 2 2,5+ j 2,5] Then the complex currents are: I 1 = 1 I 2 = 2 j2,5 =7,5+ =1,07 + j 0,36=1,13 e j18,6 [ A ] 7 j 2,5 =0,5+ =0,07+ j0,36=0,37e j 79 [ A ] 7 7 / 7

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