Electric Circuits I. Inductors. Dr. Firas Obeidat

Transcription

1 Electric Circuits I Inductors Dr. Firas Obeidat 1

2 Inductors An inductor is a passive element designed to store energy in its magnetic field. They are used in power supplies, transformers, radios, TVs, radars, and electric motors. An inductor consists of a coil of conducting wire. If current is allowed to pass through an inductor, it is found that the voltage across the inductor is directly proportional to the time rate of change of the current. = where L is the inductance of the inductor is the henry (H). There is no voltage across an inductor carrying a constant current, regardless of the magnitude of this current. Accordingly, we may view an inductor as a short circuit to dc. A sudden or discontinuous change in the current must be associated with an infinite voltage across the inductor. 2

3 Inductors Inductance is the property whereby an inductor exhibits opposition to the change of current flowing through it, measured in henrys (H). The inductance of an inductor depends on its physical dimension and construction. = where N: is the number of turns, l: is the axial length, A: is the cross-sectional area, μ: is the permeability of the air, μ r : is the relative permeability. μ= μ o μ r μ o =4π 10-7 H/m Example: Find the inductance of the air-core coil of shown figure. μ= μ o μ r =1 4π 10-7 = 4π 10-7 H/m = π = π ( ) 2 = m 2 = = 4π =1.58 μh. 3

4 Inductors Example: Given the waveform of the current in a 3 H inductor as shown in the figure, determine the inductor voltage and sketch it. a. - to -1s: since the current is zero, the voltage is zero in this interval. = = = b. -1s to 0s: = ( ) ( ( ) = c. 0s to 2s: = ( ) ( ) = d. -2s to 3s: = ( ) ( ) = d. -2s to 3s: = ( ) ( ) = 4

5 Example: Given the waveform of the current in a 0.39 H inductor as shown in the figure, determine the inductor voltage and sketch it. a. 0s to 1ms: =.. = b. 1s to 1s: = ( ) (.. ) = 5

6 Inductors-Integral Voltage-Current Relationships = = vdt Integrating the above equation gives ( ) ( ) = vdt i(t)= vdt +i(t o ) Inductors-Energy Storage = = ( )i ( ) w= = i di ( ) w(t)-w(t o )= [( ) ] If we assume that a value of t 0 is selected at which the current is zero; it is also assumed that the energy is zero at this time. We then have w= 6

7 Example: The current through a 0.1H inductor is i(t)=10te -5t, Find the voltage across the inductor and the energy stored in it. = =0.1 (10te 5t) =e 5t +t( 5)e 5t =e 5t (1-5t) V w= = (. )(10te 5t ) = (. )(100t 2 e 10t )= 5t 2 e 10t J Example: Find the current through a 5-H inductor if the voltage across it is. = > < Also, find the energy stored at t=5s, assume i>0. i(t)= vdt +i(to) = 30t 2 dt +0=6 =2t 3 A w= ( = 2 = kj 7

8 Example: The voltage across a 2 H inductor is known to be 6cos 5t V. Determine the resulting inductor current if i(t = π/2)=1 A. i(t)= 6cost dt +i( ) = ( )sin5t ( )sin5t o + i( ) i(t)=. sin5t. sin5t o + i( ) Assume k=. sin5t o + i( ) i(t)=. sin5t+k i(t=-π/2)= -. + k=1 k=1.6. sin(5 ( π/2))+k=1 i(t)=. sin5t+1.6 8

9 Series Inductors Consider a series connection of N inductors, as shown in the fig.(a), with the equivalent circuit shown in fig.(b). The inductors have the same current through them. Applying KVL to the loop, The equivalent inductance of series-connected inductors is the sum of the individual inductances. 9

10 Parallel Inductors The combination of a number of parallel inductors is accomplished by writing the single nodal equation for the original circuit, shown in fig.(a) For the special case of two inductors in parallel The equivalent inductance of parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances. 10

11 Example: The voltage across a 2 H inductor is known to be 6cos 5t V. Determine the resulting inductor current if i(t = π/2)=1 A. i(t)= 6cost dt +i(to) = ( )sin5t ( )sin5t o + i(to) i(t)=. sin5t. sin5t o + i(to) Assume k=. sin5t o + i(to) i(t)=. sin5t+k i(t=-π/2)= -. + k=1 k=1.6. sin(5 ( π/2))+k=1 i(t)=. sin5t

12 (a): (b): Example: Find the equivalent inductance of the circuit in the shown figure. The 10H, 12H, and 20H inductors are in series; thus, combining them gives a 42H inductance. This 42H inductor is in parallel with the 7H inductor so that they are combined, to give + = This 6H inductor is in series with the 4H and 8H inductors. Hence, L eq =4+6+8=18H Example: for the circuit in the shown figure, i(t)=4(2-e -10t )ma. If i 2 (0)=-1mA. Find: (a) i 1 (0), (b) v(t), (c) v 1 (t), (d)v 2 (t), (e) i 1 (t), (f)i 2 (t). i(t)=4(2-e-10t)ma, i(0)=4(2-1)=4ma i 1 (0)=i(0)-i 2 (0)=4-(-1)=5mA L eq =2+4 12=2+3=5A 12

13 v(t)=leq = e 10t = 200 e 10t (c): v 1 (t)=2 = e 10t = 80e 10t (d): Since v=v 1 +v 2 v 2 (t)=v(t)-v 1 (t)=120e 10t mv (e): i 1 (t )= + = e 10t + (f): = 3e 10t + = e 10t =8 e 10t ma i 2 (t )= + = e 10t = e 10t = e 10t +1 1 = e 10t ma 13

14 Properties of an Inductors The voltage across an inductor is zero when the current is constant. Thus, An inductor acts like a short circuit to dc. The current through an inductor cannot change instantaneously. The ideal inductor does not dissipate energy. The energy stored in it can be retrieved at a later time. The inductor takes power from the circuit when storing energy and delivers power to the circuit when returning previously stored energy. Nonideal inductor has a significant resistive component called the winding resistance. The nonideal inductor also has a winding capacitance, as shown in the figure. 14

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