Lecture 03: Review of homework problemsVijay Singh∗January 25, 2003AbstractWe will solve a couple of homework problems.Problem 7.19In the circuit of Figure 1, find the value of L such that i(t) is in phase with vs(t).+ -vs(t) = 24cos(377t + 60o)L2 ohms1326 FiFigure 1: Circuit for problem 7-19.Let I and Vsbe the phasors that correspond to i(t) and vs(t).In the frequency domain we can writeI =VsR + (jωL +1jωC)For I and Vsto be in phase, R + (jωL +1jωC) must be a real number. In other words,jωL +1jωCmust be equal to zero, that is, L =1ω2C.Thus,L =1ω2C=13772× 1326 ×10−6H.∗Professor and Chairman, Department of Electrical & Computer Engineering, University of Kentucky, Lex-ington, KY, USA. E-mail: [email protected] Lecture presented on January 24, 2003. Typeset in LATEX1Problem 7.33In the circuit of Figure 2, find vo(t). Also, using a phasor diagram, show that i1(t) + i2(t) =iS(t).is(t)20 ohms6 mH3.33 micro F10 ohmsvo(t)+-Figure 2: Time domain circuit.The phasor circuit corresponding to the circuit of Figure 2 is shown in Figure 3.Is20 ohmsj104x6x103 ohms1/(jwC) =-j30 ohms10 ohmsVo+-Figure 3: Frequency domain circuit.Vo= [(20 + j60)k(10 − j30)] × Is=(20 + j60)(10 − j30)(20 + j60) + (10 − j30)× Is=200 − j600 + j600 + 180030(1 + j)× Is=2000301 − j1 + j= Is1003(1 − j)= 0.36(−135◦) × 33.33 × (1√2− j1√2) ×√2= 0.36(−135◦) × 33.33√26(−45◦)= 10√26(−180◦)= 14.146(−180◦)So,vo(t) = 14.14 cos(104t − 180◦) V.Finding I1and I2:2I1=Vo20 + j60=14.146(−180◦)20(1 + j3)=0.7076(−180◦)√106(71.57◦)=0.7076(−180◦)3.166(71.57◦)= 0.2246(−251.6◦)I1=Vo10 − j30=14.146(−180◦)10(1 − j3)=14.146(−180◦)10 ×√106(−71.57◦)=1.4146(−180◦)3.166(−71.57◦)= 0.4486(−10.8.4◦)The phasor diagram showing IS= I1+ I2is shown in Figure 4.I1ISI2-108.4o-251.6oFigure 4: Phasor
View Full Document
Unlocking...