Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14EE201 Lecture 15 P. 1Inductors- A new circuit element ABi (t)A time-varying current through the conductor induces a voltage vAB(t) between ends of wire. vAB(t) =L di(t)dt(1)Where the proportionality constant L is the inductance of the wire. Symbol: iL(t)+ _Lvl(t)EE201 Lecture 15 P. 2Units: Henry(H) 1 H = 1 vL(t)+_Volt-sec ampInductors are energy storage elements.Example: Calculate vL(t) when iL(t)=sin(20t + /3)0.5Hi L(t)Using Eq (1),vL(t) = L diL(t) dtEE201 Lecture 15 P. 3 vL(t) = (0.5) d [ sin (20t + /3)] dt vL(t) = 10 cos (20t + /3)The integral relation is:iL(t) = -tvL() dWhere iL(-) = 0 iL(t) = 1L-t0vL() d + t0tvL() d(2)Memory1LEE201 Lecture 15 P. 4Memory: initial current flowing through inductor before initial time of interest.Example: Find iout(t) given the current profile supplied by the input source shown.+0.5His(t)+_vL1+_iout(t)0.25H2vL1123 -11t(s)is(t)Current supplied by current sourceEE201 Lecture 15 P. 5Step1: Find voltage generated by dependent source.vL1(t) = 0.5dis(t)dt2 vL1 =dis(t)dt(Slope of is – t curve)12 3 -112vL1t (sec)EE201 Lecture 15 P. 6Step 2: Find iout(t).Using integral relationship, iout(t) = iout(0) + 1L0t2vL1() d()=0=Area under vL1-t curve.1234-2-4 -6 -8t (sec)iL(t)EE201 Lecture 15 P. 7Continuity Property of InductorsThe current through an inductor is continuous even if the voltage across the inductor is discontinuous, provided the voltage is not pulsed.1HvL(t)iL(t)1-1123t (sec)vL(t)DiscontinuousvL(1-) vL(1+)+_iL(t) = 1Lt0tvL() d1 2iL(t)ContinuousiL(t-) = iL(t+) for all tEE201 Lecture 15 P. 8 Power and Energy Instantaneous power absorbed by inductor pL(t) = iL(t) vL(t) = L iL(t)diL(t)dtEnergy stored over an interval [t0, t1] is: WL (t0,t1) = t0t1p() d()t0t1iL()diL()ddiL(t0)iL(t1)iLdiLWL (t0,t1) = LWL (t0,t1) = LEE201 Lecture 15 P. 9WL(t0 , t1) = 12LiL2(t1) - 12LiL2(t0)(3)An instantaneous energy may be defined as,WL(t) = 12L iL2(t)(4)where t0 in Eq.(3) is t0= - and iL(-) = 0Inductors in Series+_-L1iin(t)+v1+-v2+-v3LeqL2L3vin(t)vin = v1 + v2 + v3 = L1diindt+ L2 diindt+ L3diindtvin = (L1 +L2 + L3)diindt Therefore, Leq = L1 + L2+ L3In series, inductors add to form equivalent inductance.EE201 Lecture 15 P. 10Inductors in Parallel+-vin(t)iL1L1iL2L2Leqiin(t)What is Leq?L1 : vin(t) = L1 diL1(t)dtdiL1dt= 1L1vin(t)L2 : vin(t) = L2 diL2(t)dtdiL2dt=1L2vin(t)EE201 Lecture 15 P. 11Equivalent circuit is:+-vin(t)Leqiin(t)Where vin(t) = Leqdiin(t)dt(5)From KCL, iin(t) = iL1(t) + iL2(t)diin(t)dt= dtdtdiL2(t)diL1(t)+vin(t)Leq=vin(t)vin(t)L2L1+EE201 Lecture 15 P. 12Leq1=11L2L1+Leq=11L1+ 1L2=L1 L2L1 + L2 Inductors in parallel behave like resistors in parallel.Current Division Formula:-two inductors in parallel-iL(-) = 0iin(t) = iL1(t) + iL2(t)EE201 Lecture 15 P. 13L1L2L2L1+iL1(t) = 1L1-tvin() d()From Eq (5),iL1(t) = 1L1-t)(diin()ddiL1(t) = L2+L2L1iin(t)=1L1iin(t)11L2L1+In general, for n parallel inductors,iLj = 1Lj 1Lnniin(t)EE201 Lecture 15 P. 14Example: Find the equivalent inductanceLeq18mH15mH85mH26mH21mHLeq=(18mH) + 21mH || 126mH= 0.018 + (0.021) (0.126)0.021+0.126= 0.018 + 11/0.0211/0.126+Leq= 0.0356 H =
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