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Electric Circuits II
The Step response of a Parallel RLC Circuit
Lecture #7
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The material to be covered in this lecture is as follows:
o The step response of a parallel RLC circuit
o The general solutions of the second order differential equation
o The form of the step Response of A Parallel RLC circuit
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After finishing this lecture you should be able to:
 Understand The step response of a parallel RLC circuit
 Solve for circuit with second order differential equation
 Know The forms of the step Response of A Parallel RLC circuit
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The step response of a parallel RLC circuit.
iL
ic
I
iR
+
R
v
t=0
C
L
DC
source
_
Circuit for Step Response of Parallel RLC
There may be or may no be energy stored in the circuit when the DC current source is applied.
Assume that the energy stored initially is zero for the step response
By KCL
iL  iC  iR  I
8.3.1
iL  v  C dv  I
R
dt
8.3.2
or
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di
since the voltage across the inductor is v  L L
dt
8.3.3
then
dv  L d 2iL
dt
dt 2
8.3.4 so
d 2iL
L
di
iL 
 CL 2  I
R dt
dt
2
d iL 1 diL iL


 I
2
RC dt LC LC
dt
8.3.5
8.3.6
Two approaches could be followed to solve for (8.3.6)
The Indirect Approach:
We can solve for iL indirectly by first finding v thus from 8.3.1
t
iL  iC  iR  I we get 1  vdt  v  C dv  I
L
R
dt
8.3.7
0
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differentiating both sides of 8.3.7
v  1 dv  C d 2v  0
L R dt
dt 2
8.3.8
three possible solutions depending on the roots of 8.3.8
st
st
v  A1e 1  A2e 2
v  B1e t cosd t  B2e t sin d t
v  D1te t  D2e t
8.3.9
to find iL for the three possible solutions
st
st
iL  I  A'1e 1  A'2e 2
iL  I  B'1et cosd t  B'2et sind t
8.3.10
iL  I  D'1tet  D'2et
where A’1 , A’2,B’1,B’2, D’1 and D’2 are arbitrary constants. This approach is cumbersome.
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The direct approach
It is easier to find the primed constants directly in terms of initials conditions or values of the response
dil
0 .
function. We would find the initial iL 0 ,

dt 

The solution of a second order differential equation with a constant forcing function = forced response
plus a response function identical in form to the natural response.
We can always write the solution for the step response in the form of
i  I forced + {function of the same form as natural response}
OR
v  V forced + {function of the same form as natural response}
Where I f and V f are the final values of the response function, see Example 8.6, 8.7, 8.8, 8.10
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Example Drill 8.6
In the circuit shown R  500, L  0.64H , C  I  F , I0  0.5 A,V0  40v and I  1 A . Find a) iR  0  b) ic  0  c)
diL  0 

 d) S
, S e) i t  fort  0
L
1 2
dt
f) v (t), for t  0 
iL
ic
iR
+
R
v
t=0
I
C
L
Vo
I0
_
Parallel RLC circuit for Example Drill 8.6
Solution:
a) iR
40 0.08 A
0   500
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 
 
 
b) iC  0   I  iR  0   iL  0   1 0.08  0.5  1.58 A






vL (0 )  vR (0 )  40  L
c)
diL (0 )
 40
dt
diL 
0 ?
dt  
diL (0 ) 1
 v0  1  40  62.5 A /sec
L
0.64
dt
d)
S1,2     2  02

2RC
 1
1
 1000,
2 500106
1
02  1 
 1.5625106

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LC 0.6410
0 1250rd /sec
d  02  2  750rd /sec
 2 106  02 s1 and s2 are complex
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S1     2 02  1000  j750
S2     2 02  1000  j750
iL (t )  I f  B1e t cosd t  B2e t sin d t
Using the given initial information if=-1A
e)
iL (0 )  0.5  I f  B1  1 B1  B1  1.5
diL (0 )
  B1  d B2  1000(1.5)  750B2  62.5
dt
1500  62.5 


  2.08333
B2  
750
iL (t )  1 e1000t (1.5cos750t  2.0833et sin750t) A
For t  0
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L
diL (t )


 0.64 1000e1000t 1.5cos750t  2.0833sin750t   

 
dt



0.64 e1000t 1.5 750(1)sin750t  2.0833 750cos750t  


 
v(t )  e1000t 40cos750t  2053.33sin750t V for t  0


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