ECE 231 -8

ECE 231 -8 - ECE-231 Circuits and Systems I Spring 2011...

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ECE-231 Circuits and Systems I Spring 2011 Session 8 Professor Stewart Personick Office: ECEC Room 321 [email protected]
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“And Now For Something Really Amazing!” Rocky and Bullwinkle, ca. 1960 Inductors
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Network Component: Inductor Wire carrying current i Magnetic field intensity: H= B/ μ 0 = Magnetic flux density/ μ 0 The line integral of: H dl around a closed path is equal to the current i passing through the closed path (Ampere’s Law) Vacuum B/ μ 0
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Network Component: Inductor Wire carrying current I The magnetic field around a current-carrying conductor contains energy. If you start with the current = 0 A, and increase it, the wire will produce a voltage that opposes the increase in the current. This voltage is given by the formula Vab = L d[i(t)]/dt; where L is the inductance of the wire between node a and node b. L has the units of Vacuum B/ μ 0 a b
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Network Component: Inductor Wire carrying current I Vacuum B/ μ 0 a b The inductance (Henries) of a straight wire is usually negligible (but not always). Therefore we usually ignore it. [Just as we usually ignore the capacitance of an open circuit]
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Network Component: Inductor Wire loop carrying current I Vacuum
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Network Component: Inductor Coil of wire carrying current I Vacuum
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Network Component: Inductor Wire carrying current I Vacuum The inductance (Henries) of a coil of wire is usually not negligible. It increases with the number of turns (but not linearly). You can find “calculators” on line that can be used to calculate the inductance of a coil having a specified diameter, a specified spacing between the turns, and a specified number of turns. Example: If the coil diameter is 1.75 inches, the number of turns is 42, and the spacing between turns is 1/8th of an inch… then the inductance of the coil is 21.5 μ H
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Network component: Inductor + - Vs= V Rs i(t) - + Vab For times t < 0 seconds, the switch is open, and i(t) = 0 Since i(t) = 0 , there is no energy stored in the magnetic field of the inductor . The switch is open for: t<0s L (Henries)
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Network component: Inductor + - Vs= V Rs i(t) - + Vab At t=0+, i(t) = 0 In the steady state (for t>> L/Rs seconds): i(t) = V/Rs and Vab(t) =0. During the transition Vab(t) = L d[i(t)]/dt Close the switch at: t=0s L (Henries)
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Network component: Inductor + - Vs= V Rs i(t) - + Vab The energy stored in the magnetic field of the inductor, in the steady state is: the integral ( from t=0 to t>> L/Rs seconds ) of i(t) x vab(t) dt = the integral ( from t=0 to t>> L/Rs seconds ) of i(t) x {L d[i(t)]/dt} dt = the integral ( from i(t)=0 to i(t)=V/Rs Amperes ) of i(t) x L di(t) = ½ x L x i2 (Joules) Close the switch at: t=0s L (Henries)
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Network component: Inductor + - Vs= V Rs i(t) - + Vab More generally: the energy stored in the magnetic field of an inductor, at any time t (seconds), is: E(t) = ½ L x [i(t)]2 (Joules) Close the switch at: t=0s L (Henries)
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Network component: Inductor + - Vs= V R s i(t ) - + Vab L KVL: -V + Rs x i(t) + L d[i(t)]/dt =0, for t>0 Therefore: i(t) + L/ Rs d[i(t)]/dt = V/ Rs , for t>0 Close the switch at: t=0s
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Network component: Inductor + - Vs= V i(t) - + Vab i(t) = V/Rs [1 - e-t/ τ ] where τ = L /Rs i=V/R s t L KVL: -V + Rs x i(t) + L d[i(t)]/dt =0, for t>0 Therefore: i(t) + L/ Rs d[i(t)]/dt = V/ Rs , for t>0 Close the switch at: t=0s R s
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It’s time for: the  Professor’s Basement Kids: you can do this at home if you have the parts in your basement
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This note was uploaded on 03/26/2012 for the course ECE 231 taught by Professor Pietrucha during the Spring '08 term at NJIT.

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ECE 231 -8 - ECE-231 Circuits and Systems I Spring 2011...

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