Electric Circuits IInductors

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Dr. Firas Obeidat

Dr. Firas Obeidat – Philadelphia University 2

Inductors

An inductor is a passive element designed to store energy in itsmagnetic field.

They are used in power supplies, transformers, radios, TVs, radars, andelectric motors.

An inductor consists of a coil of conducting wire.

If current is allowed to pass through an inductor, it is found that thevoltage across the inductor is directly proportional to the time rate ofchange of the current.

� = ���

��

where L is the inductance of theinductor 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 aninductor as a short circuit to dc.

A sudden or discontinuous change in the current must be associated withan infinite voltage across the inductor.

Dr. Firas Obeidat – Philadelphia University 3

Inductors

The inductance of an inductor depends on its physical dimension andconstruction.

� =����

�whereN: is the number of turns,l: is the axial length,A: is the cross-sectional area,μ: is the permeability of the air,μr: is the relative permeability.

μ= μo μr

μo=4π×10-7 H/m

Inductance is the property whereby an inductor exhibits opposition to thechange of current flowing through it, measured in henrys (H).

Example: Find the inductance of the air-core coil of shown figure.

μ= μo μr=1× 4π×10-7= 4π×10-7 H/m

� = π��= π×(2×10-3)2=12.57×10-6 m2

� =����

�=

����×4π×10−7×12.57×10−6

�.�=1.58 μH

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InductorsExample: Given the waveform of the current ina 3 H inductor as shown in the figure,determine the inductor voltage and sketch it.

a. -∞ to -1s: since the current is zero, thevoltage is zero in this interval.

� = ���

��

� = ��

��= �

b. -1s to 0s: � = �(� − �)

(� − (−�)= ��

c. 0s to 2s: � = �(� − �)

(� − �)= ��

d. -2s to 3s: � = �(� − �)

(� − �)= −��

d. -2s to 3s: � = �(� − �)

(∞ − �)= ��

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Example: Given the waveform of the current ina 0.39 H inductor as shown in the figure,determine the inductor voltage and sketch it.

a. 0s to 1ms: � = �.��� − �

�.��� − �= ����

b. 1s to 1s: � = �(� − �)

(�.��� − �.���)= −∞ �

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Inductors-Energy Storage

� = �� = (���

��)i

w=∫ ���′�

��

= � ∫ i′di′ �(�)

�(��)

w(t)-w(to)=�

�� [(� �) � − � ��

�]

If we assume that a value of t0 is selectedat which the current is zero; it is alsoassumed that the energy is zero at thistime. We then have

w=�

����

Inductors-Integral Voltage-Current Relationships

� = ���

��

�� =�

�vdt

Integrating the above equation gives

∫ ��′�(�)

�(��)=

�

�∫ vdt′

�

��

i(t)=�

�∫ vdt′

�

��

+i(to)

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Example: The current through a 0.1H inductor is i(t)=10te-5t, Find thevoltage 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)� =

�

�(�.�)(100t2e−10t)= 5t2e−10t J

Example: Find the current through a 5-H inductor if the voltage across itis.

� � = ����� � > �� � < � Also, find the energy stored at t=5s, assume i>0.

i(t)=�

� ∫ vdt′

�

��

+i(to) =�

�∫ 30t2dt′

�

�+0=6×

��

�=2t3 A

w=�

��(� � � − � � � =

�

�� 2×�� � − � = 156.25 kJ

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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−

�

�(�

�)sin5to+ i(��)

i(t)= �.�sin5t−�.�sin5to+ i(��)

Assume k= −�.�sin5to+ i(��)

i(t)= �.�sin5t+k

i(t=-π/2)= �.�sin(5 × (−π/2))+k=1

-�.� + k=1

k=1.6

i(t)= �.�sin5t+1.6

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Series InductorsConsider a series connection of N inductors, as shown in the fig.(a), withthe equivalent circuit shown in fig.(b). The inductors have the samecurrent through them. Applying KVL to the loop,

The equivalent inductance of series-connectedinductors is the sum of the individualinductances.

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Parallel InductorsThe combination of a number of parallel inductors is accomplished bywriting 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 ofthe sum of the reciprocals of the individual inductances.

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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−

�

�(�

�)sin5to+ i(to)

i(t)= �.�sin5t−�.�sin5to+ i(to)

Assume k= −�.�sin5to+ i(to)

i(t)= �.�sin5t+k

i(t=-π/2)= �.�sin(5 × (−π/2))+k=1

-�.� + k=1

k=1.6

i(t)= �.�sin5t+1.6

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Example: Find the equivalent inductance of the circuit in the shownfigure.

The 10H, 12H, and 20H inductors are in series; thus,combining them gives a 42H inductance. This 42Hinductor is in parallel with the 7H inductor so thatthey are combined, to give

� × ��

� + ��= ��

This 6H inductor is in series with the 4H and 8H inductors. Hence,

Leq=4+6+8=18H

Example: for the circuit in the shown figure, i(t)=4(2-e-10t)mA. If i2(0)=-1mA.Find: (a) i1(0), (b) v(t), (c) v1(t), (d)v2(t), (e) i1(t), (f)i2(t).

i(t)=4(2-e-10t)mA, i(0)=4(2-1)=4mA

i1(0)=i(0)-i2(0)=4-(-1)=5mA

(a):

(b):

Leq=2+4||12=2+3=5A

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v(t)=Leq��

��= � � −� −�� e−10t� = 200 e−10t��

v1(t)=2��

��= � � −� −�� e−10t� = 80e−10t��

Since v=v1+v2

v2(t)=v(t)-v1(t)=120e−10tmV

(d):

(c):

(e):

i1(t )=�

�∫ ���� + �� � =

�

�

���

�∫ e−10t��� + ���

�

�

=−3e−10t����

+ ��� = −�e−10t�+3�+5�=8−�e−10tmA

(f):

i2(t )=�

��∫ ���� + �� � =

�

�

���

��∫ e−10t��� − ���

�

�

=− e−10t����

− ��� = − e−10t�+1�−1�=−e−10tmA

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Properties of an Inductors

The voltage across an inductor is zero when the current isconstant. Thus, An inductor acts like a short circuit to dc.

The current through an inductor cannot changeinstantaneously.

The ideal inductor does not dissipate energy. The energy storedin it can be retrieved at a later time. The inductor takes powerfrom the circuit when storing energy and delivers power to thecircuit when returning previously stored energy.

Nonideal inductor has a significant resistive component calledthe winding resistance. The nonideal inductor also has awinding capacitance, as shown in the figure.

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