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Parallel RLC Network
Objective of Lecture
 Derive the equations that relate the voltages across and the
currents flowing through a resistor, an inductor, and a
capacitor in parallel as:
 the unit step function associated with voltage or current
source changes from 1 to 0 or
 a switch disconnects a voltage or current source in the circuit.
 Describe the solution to the 2nd order equations when the
condition is:
 Overdamped
 Critically Damped
 Underdamped
RLC Network
 A parallel RLC network where the current source is
switched out of the circuit at t = to.
Boundary Conditions
 You must determine the initial condition of the
inductor and capacitor at t < to and then find the final
conditions at t = ∞s. Replace the capacitor with an
open circuit and the inductor with a short circuit.
 Since the current source has a magnitude of Is at t < to
 iL(to-) = Is and v(to-) = vC(to-) = 0V
 vL(to-) = 0V and iC(to-) = 0A
 Once the steady state is reached after the current source
been removed from the circuit at t > to and the stored
energy has dissipated through R.


iL(∞s) = 0A and v(∞s) = vC(∞s) = 0V
vL(∞s) = 0V and iC(∞s) = 0A
Selection of Parameter
 Initial Conditions
 iL(to-) = Is and v(to-) = vC(to-) = 0V
 vL(to-) = 0V and iC(to-) = 0A
 Final Conditions
 iL(∞s) = 0A and v(∞s) = vC(∞s) = oV
 vL(∞s) = 0V and iC(∞s) = 0A
 Since the current through the inductor is the only
parameter that has a non-zero boundary condition,
the first set of solutions will be for iL(t).
Kirchoff’s Current Law
iR (t )  iL (t )  iC (t )  0
v(t )  vR (t )  vL (t )  vC (t )
dvC (t )
vR (t )
 iL (t )  C
0
R
dt
diL (t )
vL (t )  v(t )  L
dt
d 2iL (t ) L diL (t )
LC

 iL (t )  0
2
dt
R dt
d 2iL (t ) 1 diL (t ) iL (t )


0
2
dt
RC dt
LC
General Solution
1
1
s 
s
0
RC
LC
2
2
1
1
 1 
s1  
 
 
2 RC
 2 RC  LC
2
1
1
 1 
s2  
 
 
2 RC
 2 RC  LC
s1      
2
2
o
s 2      
2
1

2 RC
1
o 
LC
2
o
s  2s    0
2
2
o
Note that the equation for the natural frequency of the RLC
circuit is the same whether the components are in series or in
parallel.
Overdamped Case
  > o
 implies that L > 4R2C
s1 and s2 are negative and real numbers
iL1 (t )  A1e
s1t
iL 2 (t )  A2 e
s2 t
t  t  to
iL (t )  iL1 (t )  iL 2 (t )  A1e
s1t
 A2 e
s2 t
Critically Damped Case
   o
 implies that L = 4R2C
s1 = s2 = -  = -1/2RC
iL (t )  A1e
t
 A2 te
t
Underdamped Case
  < o
 implies that L < 4R2C
s1     2  o2    jd
s2          jd
2
2
o
d  o2   2
iL (t )  e t [ A1 cos d t  A2 sin d t ]
Solve for Coefficients A1 and A2
 Use the boundary conditions at to- and t = ∞s to solve
for A1 and A2.

iL (to )  I S
 Since the current through an inductor must be a
continuous function of time.




iL (to )  iL (to )  iL1 (to )  iL 2 (to )  I S
A1e s1 0 s   A2 e s2 0 s   A1  A2  I S
 Also know that
diL (to ) d
vL (to )  L
 iL1 (to )  iL 2 (to )  0
dt
dt
s1 A1e s1 0 s   s2 A2 e s2 0 s   s1 A1  s2 A2  0
Other Voltages and Currents
 Once current through the inductor is known:
diL (t )
vL (t )  L
dt
vL (t )  vC (t )  vR (t )
dvC (t )
iC (t )  C
dt
iR (t )  vR (t ) / R
Summary
 The set of solutions when t > to for the current through the
inductor in a RLC network in parallel was obtained.
 Selection of equations is determine by comparing the natural
frequency o to .
 Coefficients are found by evaluating the equation and its first
derivation at t = to-.
 The current through the inductor is equal to the initial
condition when t < to
 Using the relationships between current and voltage, the
voltage across the inductor and the voltages and currents
for the capacitor and resistor can be calculated.