Electric CircuitsGeneral & Particular Solutions

Vineet Sahula

sahula@ieee.org

First & Higher Order Differential Equations – p. 1/13

Linear Diff. Eq.

a0di(t)

dt+ a1i(t) = v(t)

a0d

ni

dtn+ a1

dn−1

i

dtn−1+ ... + an−1

di

dt+ ani = v(t)

v(t) is forcing function or excitation

First & Higher Order Differential Equations – p. 2/13

Integrating Factor

di

dt+ Pi = Q

Using Integrating Factor (I.F.) ePt we get

ePtdi

dt+ PiePt = QePt

d

dt(iePt) = QePt

Solving leads to

iePt =

Z

QePtdt + K

⇒ i = e−Pt

Z

QePtdt + Ke−Pt

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

Pdt

First & Higher Order Differential Equations – p. 3/13

Network solution

I part is Particular integral & II part is Complementary function

i = e−Pt

Z

QePtdt + Ke−Pt

Q is forcing function & K is arbitrary constant

Thus, with t → ∞ i.e. Steady State

limt→∞

Ke−Pt = 0

i(∞) = limt→∞

i(t) = limt→∞e−Pt

Z

QePtdt

Whereas, with t → 0 i.e. Initial condition

i(0) = limt→0

i(t) = limt→0e−Pt

Z

QePtdt + K

In case, P & Q are constants,

i(0) =Q

P+ K = K2 + K

i(∞) =Q

P+ K = K2

In general,

i(t) = iP + iC = iss + it

First & Higher Order Differential Equations – p. 4/13

Example

For an RL circuit under switched-on condition

Ldidt

+ Ri = V i.e. didt

+ RL

i = VL

with P = RL

&Q = VL

i = e−Pt∫

QePtdt + Ke−Pt

→ i = VR

+ Ke−Rt

L

In general, when P & Q are constants, i = K2 + K1e−

t

T

In case, P &Q are constants,

K2 = i(∞)

K2 + K1 = i(0)

⇒ K1 = i(0) − i(∞)

⇒ i = i(∞) − [i(∞) − i(0)]e−t

T

First & Higher Order Differential Equations – p. 5/13

Example-2

L R

R

K1

2

Vi

Determine current when K is CLOSED at t = 0 and later after

steady state is reached when K is OPENED

at t = 0 i(∞) = VR1

i(0) = VR1+R2

& T = LR1

⇒ i = VR1

(

1 −R1

R1+R2e−

R1t

L

)

First & Higher Order Differential Equations – p. 6/13

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 simplified by using equivalence

conditions so as to represented by a single equivalent

L/C/R

Solve many examples !!

First & Higher Order Differential Equations – p. 7/13

Initial Conditions in Networks

Resistor: VR = iR the current changes instanteneously, if the voltagechanges instanteneously

Inductor: vL = L ·diL

dt, diL

dtfor L is finite, hence current CANNOT

change instanteneously; BUT an arbitrary voltage may appearacross it

Inductor: iC = C ·dvC

dt, dvC

dtfor C is finite, hence voltage

CANNOT change instanteneously; BUT an arbitrary current mayappear across it

Element Equivalant ckt at t = 0

R R

L Open Ckt (OC)

C Short Ckt (SC)

L, I0 Current source I0 in parallel with OC

C, V0 Voltage source V0 in series with SC

First & Higher Order Differential Equations – p. 8/13

Final Conditions in Networks

Element, IC Equivalant ckt at t = ∞

R R

L Open Ckt (OC)

C Short Ckt (SC)

L, I0 Current source I0in parallel with SC

C, V0 Voltage source V0 in series with OC

First & Higher Order Differential Equations – p. 9/13

Two special cases- Initial Conditions

A loop or mesh containing a VOLTAGE source Vs with only

capacitors,

implying a virtual short-circuit across Vs;

Imagine infinite current to flow through capacitors so as to

charge them to appropriate voltages instanteneously

In a dual situation: A node connected with a CURRENT

source Is with only Inductors in other branches

implying a virtual open-circuit across Is;

Imagine infinite voltage across Is to exist so as to drive

finite FLUX in all the inductors to bring appropriate current

in them instanteneously

First & Higher Order Differential Equations – p. 10/13

Second Order Diff. Equations

a0d2i

dt2+ a1

di

dt+ 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

a0m2kemt + a1mkemt + a2kemt = 0

Charateristic Equation, a0m2 + a1mk + a2k = 0

m1, m2 = −a1

2a0±

12a0

√

a21 − 4a0a2

i(t) = k1em1t + k2e

m2t

First & Higher Order Differential Equations – p. 11/13

Solving Second order Diff. Eqns.

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

Simple roots i(t) = k1em1t + k2e

m2t

Equal roots m1 = m2 = m

⇒ i(t) = k1emt + k2te

mt

Complex (conjugate) roots m1, m2 = −σ ± j · ω

i(t) = k1e−σte+jωt + k2e

−σte−jωt

i(t) = e−σt(k1e+jωt + k2e

−jωt)

i(t) = e−σt(k3 cos ωt + k4 sinωt)

i(t) = e−σtk5 cos(ωt + φ)

First & Higher Order Differential Equations – p. 12/13

Solving Second order Diff. Eqns.- Initial Conditions

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

is independent/unknown [i is independent in a series

circuit; v is independent in a parallel circuit ]

Second IC is based on first order differential of the same

parameter,di

dt(0+) or

dv

dt(0+)

First & Higher Order Differential Equations – p. 13/13