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October 5, 2012
EC - 304 Electrical Network Theory
Department of Electronics & Communication
EngineeringNorth-Eastern Hill UniversityBasic RL and RC circuits
Dr. L. Joyprakash Singh
1. The Source-Free RL-Circuit[with dc source initially]:The
analysis of circuits containing inductors and/or capacitors is
dependent upon theformulation and solution of the
integrodifferential equations that characterize the circuits.The
equation we obtain is a homogeneous linear differential equation
and is simply
a differential equation in which every term is of the first
degree in the dependent variableor one of its derivatives.The
solution of the differential equation represents a response of the
circuit, and it isknown by many names. Since this response depends
upon the general nature of the circuit(the types of elements, their
sizes, the interconnection of the elements), it is often calleda
natural response. However, any real circuit we construct cannot
store energy forever;the resistances intrinsically associated with
inductors and capacitors will eventually convertall stored energy
into heat. The response must eventually die out, and for this
reason itis frequently referred to as the transient response.
Finally, we should also be familiarwith the mathematicians
contribution to the nomenclature; they call the solution of a
homogeneous linear differential equation a complementary
function.
2. Case-I[dc source]: Transient analysis of the simple series RL
circuit. Let us designatethe time-varying current as i(t); we will
represent the value of i(t) at t = 0 as Io; in otherwords, i(0) =
Io.We therefore have
vR + vL = 0
Ri + Ldi
dt
= 0
di
dt+
R
Li = 0 (2.1)
Our goal is to obtain an expression fori(t)which satisfies the
above equation andalso has the value Io at t = 0
R L
i(t)
+
vR
+
vL
Figure 2.1: A series RL circuit for i(t) isto be determined,
subject to the initial con-
dition that i(0) = Io
Rearranging the Eq. 2.1,
di
i=
R
Ldt (2.2)
Since the current is Io and i(t) at time t, wemay equate the two
definite integrals whichare obtained by integrating each side
be-tween the corresponding limits;
i(t)
Io
di
i=
t
0
R
Ldt (2.3)
+
vS
t = 0 R
L
i(t)
vR ++
vL
Figure 2.2: A series RL circuit for i(t) isto be determined,
subject to the initial con-
dition that i(0) = IoPerforming indicated integration,
ln i|i
Io=
R
L
tt
0
(2.4)
Note:Practice helps analysis perfect!
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EC - 304 ENT Basic RL and RC circuits Dr. L. Joyprakash
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which results in
ln i ln Io = R
L(t 0)
lni
Io=
R
L t
i
Io= e
R
Lt
Therefore, we find that the current i(t) is given by
i(t) = IoeR
Lt =
Vo
Re
R
Lt (2.5)
Case-II[dc source]: Integrating both sides of Eq. (2.2) by
including a constant of inte-gration, we have
dii
=
RL
dt + K
Thus,
ln i = R
Lt + K (2.6)
The constant K cannot be evaluated by substitution of Eq. (2.6)
in the original differentialequation (2.1); identity 0 = 0 will
result, because Eq. (2.6) is a solution of Eq. (2.1) for anyvalue
of K. The constant of integration must be selected to satisfy the
initial conditioni(0) = Io., Thus, at t = 0, Eq. (2.6) becomes
ln Io = K
Using the above value of K in Eq. (2.6) to obtain the desired
response
ln i = R
Lt + ln Io
ln i ln Io = R
Lt
lni
Io=
R
Lt
i
Io= e
R
Lt
Therefore, we finally have
i = IoeR
Lt =
Vo
Re
R
Lt (2.7)
3. The InductorThe mathematical model of an ideal induc-tor is
defined by a simple differential equa-tion as
v = Ldi
dt(3.1)
We bear in mind that v and i are functionsof time, if needed, we
can emphsize this factthat by writing v(t) and i(t), instead.
Li
+ vL
Figure 3.1: Electrical symbol and current-voltage conventions
for an inductor
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EC - 304 ENT Basic RL and RC circuits Dr. L. Joyprakash
Singh
Integral Voltage-Current RelationshipsWe re-arrange the Eq. 3.1
as
di =1
Lv dt (3.2)
Integrate the above equation between the time limits to and
t:i(t)
i(to)
di =1
L
t
to
v(t) dt (3.3)
which leads to the equation
i(t) i(to) =1
L
t
to
v(t) dt (3.4)
or
i(t) =1
L
t
to
v(t) dt + i(to) (3.5)
Equation 3.5 may also be written as an indefinite integral plus
a constant of intergration:
i(t) =1
L
v dt + k (3.6)
We may also assume that we are solving a realistic problem in
which the selection ofto as insures no current or energy in the
inductor. Thus, if i(to) = i() = 0, then
i(t) =1
L
t
v dt (3.7)
Energy Storage:The power delivered to an inductor is
p = vi
= Lidi
dt(3.8)
and the energy stored in its electric field is therefore
t
to
p dt = L
t
to
idi
dtdt
= L
t
i(to)
i di
=1
2L{[i(t)]2 [i(to)]
2} (3.9)
and thus
wL(t) wL(to) =1
2L{[i(t)]2 [i(to)]
2} (3.10)
If we select a zero-energy reference at to, implying that the
inductor current is also zero at
that instant, thenwL(t) =
1
2Li2 (3.11)
ECE, NEHU, Shillong Page 3 of 3 October 5, 2012
