1 Order Transient

8/10/2019 1 Order Transient

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Electric CircuitsGeneral & Particular SolutionsVineet [email protected]

First & Hi her Order Differential E uations . 1 1


2/13

Linear Diff. Eq.

a 0di (t )

dt + a 1 i (t ) = v (t )

a 0d n i

dt n + a 1

d n 1 i

dt n 1 + ... + a n 1

di

dt + a n i = v (t )

v (t ) is forcing function or excitation



3/13

Integrating Factor

didt

+ P i = Q

Using Integrating Factor (I.F.) eP t we get

eP t didt

+ P ie P t = Qe P t

ddt

( ie P t ) = Qe P t

Solving leads to

ie P t = Z Qe P t dt + K i = e P t Z Qe P t dt + Ke P t

For P being a function of time, I.F. will be eR P dt



4/13

Network solutionI part is Particular integral & II part is Complementary function

i = e P t Z Qe P t dt + Ke P tQ is forcing function & K is arbitrary constant

Thus, with t i.e. Steady State

limt

Ke P t = 0

i(

) = limt i(t) =

lim t e P t

Z Qe P t dt

Whereas, with t 0 i.e. Initial condition

i (0) = limt 0

i ( t ) = lim t 0 e P t Z Qe P t dt + K In case, P & Q are constants,

i (0) = QP

+ K = K 2 + K

i ( ) = Q

P

+ K = K 2

In general,

i ( t ) = iP + iC = iss + i tFirst & Hi her Order Differential E uations . 4 1


5/13

Example

For an RL circuit under switched-on condition

L didt + Ri = V i.e. di

dt + RL i =

V L with P =

RL &Q =

V L

i = e

P t Qe P t dt + Ke

P t

i = V R + Ke

RtL

In general, when P & Q are constants, i = K 2 + K 1e

tT

In case, P &Q are constants,

K 2 = i( )

K 2 + K 1 = i(0) K 1 = i(0) i( ) i = i( ) [i( ) i(0)]e

tT



6/13

Example-2

L R

R

K1

2

Vi

Determine current when K is CLOSED at t = 0 and later aftersteady state is reached when K is OPENED

at t = 0 i( ) = V R 1

i(0) = V R 1 + R 2 & T = LR 1

i = V R 1 1 R 1R 1 + R 2 e

R 1 t

L



7/13

More Complicated Networks

Networks described by one time-constant ?

Simple circuits having simple RC or RL combinations

Containg single L or C, but in combination of any number of resistors, R

Networks, which can be simplied by using equivalence

conditions so as to represented by a single equivalent L/C/R

Solve many examples !!



8/13

Initial Conditions in NetworksResistor: V R = iR the current changes instanteneously , if the voltagechanges instanteneously

Inductor: vL = L di Ldt

, di Ldt

for L is nite, hence current CANNOTchange instanteneously ; BUT an arbitrary voltage may appear across it

Inductor: iC = C dv Cdt , dv C

dt for C is nite, hence voltage CANNOTchange instanteneously ; BUT an arbitrary current may appear across it

Element Equivalant ckt at t = 0

R R

L Open Ckt (OC)

C Short Ckt (SC)

L, I 0 Current source I 0 in parallel with OC

C, V 0 Voltage source V 0 in series with SC



9/13

Final Conditions in Networks

Element, IC Equivalant ckt at t =

R R

L Open Ckt (OC)

C Short Ckt (SC)

L, I 0 Current source I 0in parallel with SCC, V 0 Voltage source V 0 in series with OC



10/13

Two special cases- Initial Conditions

A loop or mesh containing a VOLTAGE source V s with onlycapacitors,

implying a virtual short-circuit across V s ;Imagine innite current to ow through capacitors so as to

charge them to appropriate voltages instanteneously

In a dual situation: A node connected with a CURRENT source I swith only Inductors in other branches

implying a virtual open-circuit across I s ;

Imagine innite voltage across I s to exist so as to drive niteFLUX in all the inductors to bring appropriate current in them

instanteneously



11/13

Second Order Diff. Equations

a 0d2idt 2

+ a1didt

+ a2i = v(t )

To satisfy the equation, the solution function MUST be of such form

that all three terms are of SAME form.

i(t ) = kemt

a 0m2ke

mt+ a1mke

mt+ a2ke

mt= 0

Charateristic Equation, a0m 2 + a1mk + a2k = 0

m 1, m 2 = a 12a 0 12a 0

a 21 4a 0a 2

i(t ) = k1em 1 t + k2em 2 t



12/13

Solving Second order Diff. Eqns.

roots may simple (real), equal OR complex (conjugate)

Simple roots i(t ) = k1em 1 t + k2em 2 t

Equal roots m 1 = m 2 = m i(t ) = k1emt + k2te mt

Complex (conjugate) roots m 1, m 2 = j

i(t ) = k1e

t e+ jt + k2e

t e

jt i(t ) = e t (k1e+ jt + k2e jt ) i(t ) = e t (k3 cos t + k4 sin t )

i(t ) = e t

k5 cos(t + )


g


13/13

o v ng econ or er . qns.-Initial Conditions (IC)

Two constatns k1 & k2 need be evaluated

This requires two IC to be formulated

First IC is computed as either i(0+) OR v(0+) , whichever isindependent/unknown [ i is independent in a series circuit; v isindependent in a parallel circuit ]

Second IC is based on rst order differential of the sameparameter,

didt

(0+) or dvdt

(0+)

First & Hi her Order Differential E uations 13 1

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