《电路》（英文版）14-7Applications in circuit analysis

8 14-7 Applications in circuit analysis We consider each term of the fourier aeries representing the voltage as a single source. The equivalent impedance of the network at na is used to compute the current at that harmonic XL(n=noL and xc(n)=-1/nac The sum of these individual responses is the total response l

§14-7Applications in circuit analysis We consider each term of the Fourier aeries representing the voltage as a single source. The equivalent impedance of the network at is used to compute the current at that harmonic. n XL(n) = nL and XC(n) = −1/ nC The sum of these individual responses is the total response i

Example: R If U=100+50sin@ t 25sin 3a tV, 十 0=5001/sR=5.=20mh. Find i u(t) Solution:At00=0,h=51m=100/5=20A Ato=5001/Z=5+j50020)10-3=5+j10 50 ∠0° 2 4.48 ∠-6349Ai()=4.48imt-63.4°)A ()1.25∠63.4°√2 At30=15001/13-5+150020)103=5+j30 5 ∠0 v(3) 2 0.823 ∠-80.5°4i(3)=0823sin(30t-80.5°)4 3) 30.4∠80.5 2 i=20+448sin(at-63.4°)+0.823sin(3Ot-80.5°)A P=VoIo, +vl cos 9+v31(3) cos 93=2053w =M24420.8232=2025AP=P2R=2053W 十 2

Solution: At , Z(0)=5 I 0 = 0 (0)=100/5=20A At =500 1/s Z(1)=5+j500(20)10-3  =5+j10 A Z V I =  −      = = • • 63.4 2 4.48 11.25 63.4 0 2 50 (1) (1) (1) i (1) = 4.48sin( t − 63.4)A At 3  =1500 1/s Z(3)=5+j1500(20)10-3=5+j30 A Z V I =  −      = = • • 80.5 2 0.823 30.4 80.5 0 2 25 (3) (3) (3) i (3) = 0.823sin(3 t − 80.5)A i = 20 + 4.48sin( t − 63.4) + 0.823sin(3 t − 80.5)A 500 1/ , 5, 20 . . 100 50sin 25sin3 , : s R L mH Find i If t t V Example = = = = + +     P =V(0) I(0) +V(1) I(1) cos1 +V(3) I(3) cos3 = 2053W         I = + + = 20.25A P = I R = 2053W 2 0.823 2 4.48 20 2 2 2 2 i R L − + (t)