Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8EE201 Lecture 33 P. 1Phasors (cont.)Example: Find the Thevenin equivalent for the circuit shown below.1 H2+_vs(t)=10cos(t)0.5voc2FStep 1: Compute Voc using phasor methods Vs = 10 0 Using nodal analysis at A.(VA-10) 1 2 j + 0.5 Voc + (VA –Voc)=0 (1)Because IL = YL(j) VL = (1/jL)(VA –Voc) vocABEE201 Lecture 33 P. 2Grouping like terms in Eq(1) (0.5 - j) VA + (0.5 + j) Voc= 5 (2)nodal analysis at B,(1/j)(Voc - VA) + j 2 Voc = 0Because Ic = Yc(j)Vc = jC Voc-j Voc + j VA + j2 Voc = 0VA = -Voc (3)Putting Eq. (3) into Eq. (2) j2 Voc = 5 Voc= -j2.5 = 2.5 -90 voc = 2.5 cos(t-90) V = 2.5 sin(t) VEE201 Lecture 33 P. 3Step 2: Find Isc+_10cos(t)0.5voc2FIsc+Voc__1 H2Since voc = 0, dependent current source is deactivated and acts as an open circuit element. The capacitor is by-passed by the short between ‘+’ & ‘-’.Equivalent impedance for resistor and inductor in series.Zeq = R + jL = 2 + j = 5 26.6EE201 Lecture 33 P. 4Isc = 10 05 26.6= 4.47 -26.6isc(t) = 4.47 cos (t - 26.6 )Step 3: Find ZTH and draw Thevenin equivalent ZTH = VocIsc= 2.5 -904.47 -26.6= 0.56 -63.4ZTH = 0.56 [ cos (-63.4) + j sin (-63.4)ZTH = 0.25 – j0.5EE201 Lecture 33 P. 50.25 – j0.5+_=+_2F0.252.5 sin(t)Example: Find the value of L so that the current is in phase with the source voltage.+_vs(t) =10cos(t + )CRis(t)L2.5 sin(t)EE201 Lecture 33 P. 6Step1: Evaluate equivalent impedance.Zeq(j) = R + jL - j1CZeq(j) = R + j(L -C1)Step 2: Relate current and voltageZeq(j) = VsIs= 10 Is There can be no imaginary component in Zeq(j). L - 1C= 0L = 12CEE201 Lecture 33 P. 7Example: Find vs(t).+vS__10013.3F133mHIs(t) = 2cos(500t)Step 1: Find equivalent impedance(admittance).Yeq(j) =1R+ jC +1jLYeq(j) = 1100+ j(500x13.3x10-6) - j1500x13.3x10-3Yeq(j) = 0.01 + j6.65x10-3 – j 1.5x10-2EE201 Lecture 33 P. 8Yeq(j) = 0.01 – j8.39x10-3 = 0.013 -40Zeq(j) = 1Yeq(j) =76.9 -40Step 2: Compute vS(t) using phasors.is(t)Is = 2 0Vs = Zeq (j) IsVs = 153.840or explicitly in timevS(t) = 153.8 cos(500t +
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