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ELE 1001: Basic Electrical Technology
Lecture 5
Inductor
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Overview of topics
 What is an Inductor?
 Transient behavior of an Inductive circuit
 Inductor as an energy storage element.
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Inductors
 Inductor is a passive electric device that stores
energy in its magnetic field when a current flows
through it
 A coil of wire wound on a core

Air core Inductor, iron core inductor
 Circuit representation is
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Inductive Circuit
 Inductance (L) : Property which opposes the rate of
change of current.
 The voltage induced in the inductor is proportional to the
rate of change of current flowing through it
 eL = L (di/dt)
i v
+
 Unit is Henry (H).
L
L
 This proportionality constant is the self inductance
or inductance (L)
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Equivalent Inductance
 In series
𝑳𝒆𝒒 = 𝑳𝟏 + 𝑳𝟐 + … … + 𝑳𝒏
 In Parallel
𝟏
𝟏
𝟏
𝟏
=
+ + …….+
𝑳𝒆𝒒 𝑳𝟏 𝑳𝟐
𝑳𝒏
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Growth of current in an Inductive Circuit
Applying KVL,
𝑽 −
𝒅𝒊
𝑳
𝒅𝒕
−𝑹𝒊=𝟎
Initial Conditions,
𝑨𝒕 𝒕 = 𝟎 𝒔𝒆𝒄, 𝒊 = 𝟎 𝑨
Final current & voltage equation,
𝒊=
𝑽
𝑹
𝑹
𝟏 −
−
𝒕
𝒆 𝑳
𝑹
𝒗𝑳 =
− 𝑳 𝒕
𝑽𝒆
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Growth of current in an Inductive Circuit
 Time Constant,  = L/R
 Time taken by the current through the inductor to reach its
final steady state value, had the initial rate of rise been
maintained constant
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Decay of current in an Inductive Circuit
 Initial current is through inductor is I0 = V/R
 At t =0, switch is moved from position a to b
t0
Applying KVL,
𝑳
𝒅𝒊
𝒅𝒕
+𝑹𝒊=𝟎
Using initial conditions and solving,
𝒊 = 𝑰𝟎 ( 𝒆
𝒗𝑳 = −𝑽 (
−
𝑹
𝑳
𝒕
)
R

a
V
b
𝑹
− 𝑳 𝒕
𝒆
)
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i
v
L
L

Decay of current in an Inductive Circuit
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Energy Stored in an Inductor
 Instantenous power ,
𝒅𝒊
𝒑 = 𝒗𝑳 . 𝒊 = 𝑳 𝒊
𝒅𝒕
 Energy absorbed in ‘dt’ time is
𝒅𝒘 = 𝑳 𝒊 𝒅𝒊
 Energy absorbed by the magnetic field when
current is increased from 0 to I amperes, is
𝑰
𝟏 𝟐
𝑾=
𝑳 𝒊 𝒅𝒊 = 𝑳 𝑰
𝟐
𝟎
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Summary
 Inductor stores energy in its magnetic field.
 When a series RL circuit is connected to a d.c voltage
source, there is an exponential growth of current
through the inductor and the current decays
exponentially when the voltage source is removed.
 Time constant of a series RL circuit is
𝐿
𝑅
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𝑠𝑒𝑐𝑜𝑛𝑑𝑠.
Illustration 1
An R-L series circuit is designed for a steady current of
250mA. A current of 120 mA flows in the circuit at an
instant 0.1 sec after connecting the supply voltage.
Calculate i) time constant of the circuit ii) the time from
closing the circuit at which the circuit current has reached
200 mA.
Ans:
(i) Time constant = 0.1529 s
(ii) t= 0.2461 s
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Illustration 2
A coil of resistance 8Ω and inductance of 0.5H is
connected to 110V d.c supply through a switch. The switch
is closed at t=0 seconds. Find
 Rate of change of current at the instant of closing of
switch.
 Final steady value of current.
 Time constant of the circuit
 Time taken by the circuit to rise to half of steady state
value of current.
Ans: (i) 220 A/s ; (ii) 13.75 A (iii) 0.0625 seconds (iv) 0.04332 seconds
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