-
)(
:
: ! Transients in DC Circuits II
#$%&'()*+#,-%.%/%Natural Response of RLC Circuit
0-%'$'1-&12.3456&$'()*+.7'$.$+&8
#,-9#$.0:;.)3=&>..A&&;7C-D
E-F)(#,-'$'1-&1I$.J./KLV0.0:;.)CNO
'&(#,-I0#$.0:;.)CN6#$.
+
-
i
iL iRC VO IO
V
E-P)(
&1I$.$Q8J4'()*+&6&R7vSTU6=C
'V&12.J&WL.6$.)/:
01 00
=+++ dtdvCIvd
LRv
t
t
(1)
01 22
=++dt
vdCLv
dtdv
R (2)
0122
=++LCv
dtdv
RCdtvd
(3)
.%/%&'IQXY.Z5%:*+WL#$%>'$#,-% Step Response of Parallel
RLC Circuit
'1V&1[&+'IQU().L'1-&124\:2] C^6._6
E-F)(E$A.%&;&.
iC
LiRi
+
-
V
-
)(
E-P)(
#$'$:V`7iL&+C%WA$.1./E,Da0KL
&.0:b.=[&`d7V..0,.[&C%.$QO.%EL:$6
+e$*PXV0U()OU6=C'$E: Iiii CRL =++ (4)
or
IdtdvC
RviL =++ (5)
dtdiLv L= (6)
2
2
dtidL
dtdv L
= (7)
Idt
idLCdtdi
RLi LLL =++ 2
2
(8)
LCLCi
dtdi
RCdtid LLL 112
2
=++ (9)
.EL:
.&$.&J/.J&.ELC-DiL(0)
diL(0)/dtEL'$E+e'f:[EF6:Q*+'()*+.
i = If + function of the same form as the natural response (10)
or
v = Vf + function of the same form as the natural response (11)
KLIfVf1/.&$..
#$%&'()C%:*+WL#,-%''$.%/% The natural responses of a series
RLC circuits
E-F.A&E$A'()5%:*+WLC-D)(&12Wh9&(0
'V./$Y&'1-:
01 00
=+++ VidCdtdiLRi
t
t
(12)
'$E+eC%:$.(EQ:
022
=++Ci
dtidL
dtdiR (13)
022
=++LCi
dtdi
LR
dtid
(14)
-
)(
I(0) V(0)+
-
i
E-P)(
'V&$.ij.4C-D:
012 =++LC
sLR
s (15)
]['V.:
LCL
RL
Rs
122
2
2,1 -
-= (16)
or 20
22,1 waa --=s (17)
NO() 'V 4C-D: srad
LR /
2=a (18)
f0O:
sradLC
/10 =w (19)
$$L.l3l&[O:
tstseAeAti 21 21)( += (overdamped) (20)
teBteBti dt
dt
ww
aa sincos)( 21 += (underdamped) (21) tt eDteDti aa -- += 21)(
(critically damped) (22)
.%/%&'IQXY.Z5%:*+WL#$%''$#,-% Step Response of Series RLC
Circuit
#,-'$&12'.(EZ6C->'$&S%EZ./6=aFO
E-F.A&&;77)(.0:;.)9`dS%E$A
Qi>O.6&(.
R LC
+ - +
+
-
-
VR VL
CVi
-
)(
E-P)(
E+e&1I$*PXV0U()'$:
CvdtdiLRiV ++= (23)
n(O6i#,-'$&12vC.3:
dtdvCi C= (24)
2
2
dtvdC
dtdi C
= (25)
LCV
LCv
dtdv
LR
dtvd CCC
=++2
2
(26)
$L.l3l&[O: tsts
fC eAeAVv 21 21 ++= (overdamped) (27) teBteBVv d
td
tfC ww
aa sincos 21--
++= (underdamped) (28)
teDteDVv tfC aa
-++= - 21 (critically damped) (29)
KLVf#,-'$&1I$.1./vCE-FC%1!6V )(&+&1[>7
V.
-
)(
Example (5): The initial energy stored in the circuit is zero.
At t = 0, a dc current source of 24 mA is applied to the circuit.
The value of the resistor is 400 .
1) What is the initial value of iL? 2) What is the initial value
of diL/dt? 3) What are the roots of the characteristic equation? 4)
What is the numerical expression for iL(t) when 0t ?
+
-
V
iC iL iR
t
-
)(
Example (6):
+
-
VC
No energy is stored in the inductor or the capacitor when the
switch is closed. Find vC(t) for 0t .
Solution: The roots of the characteristic equation are
./)48001400(
/)48001400()4.0)(1.0(10
2.0280
2.0280
2
62
1
sradjs
sradjs
--=
+-=-
+-=
The roots are complex, so the voltage response is underdamped.
Thus 0,4800sin4800cos48)( 1400214001 ++= -- tteBteBtv ttC No energy
is stored in the circuit initially, so both vC(0) and dvC(0+)/dt
are zero. Then: 1480)0( BvC +== and
12 140048000)0( BB
dtdvC
-
==
+
Solving these two equations: VB 481 -= and VB 142 -= Therefore
the solution for vC(t) is ( ) 0,4800sin144800cos4848)( 14001400 --=
-- tVtetetv ttC
