ckt transients
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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
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