Series RLC Network
Objective of Lecture
 Derive the equations that relate the voltages across a
resistor, an inductor, and a capacitor in series as:
 the unit step function associated with voltage or current
source changes from 0 to 1 or
 a switch connects a voltage or current source into the
circuit.
 Describe the solution to the 2nd order equations when
the condition is:
 Overdamped
 Critically Damped
 Underdamped
Series RLC Network
 With a step function voltage source.
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.
 Since the voltage source has a magnitude of 0V at t < to
 i(to-) = iL(to-) = 0A and vC(to-) = 0V
 vL(to-) = 0V and iC(to-) = 0A
 Once the steady state is reached after the voltage source
has a magnitude of Vs at t > to, replace the capacitor with
an open circuit and the inductor with a short circuit.


i(∞s) = iL(∞s) = 0A and vC(∞s) = Vs
vL(∞s) = 0V and iC(∞s) = 0A
Selection of Parameter
 Initial Conditions
 i(to-) = iL(to-) = 0A and vC(to-) = 0V
 vL(to-) = 0V and iC(to-) = 0A
 Final Conditions
 i(∞s) = iL(∞s) = 0A and vC(∞s) = Vs
 vL(∞s) = 0V and iC(∞s) = 0A
 Since the voltage across the capacitor is the only
parameter that has a non-zero boundary condition,
the first set of solutions will be for vC(t).
Kirchhoff’s Voltage Law
 v(t )  0
diL (t )
vC (t )  L
 RiL  VS  0
dt
dvC (t )
iC (t )  C
dt
iL (t )  iC (t )
d 2 vC (t )
dvC (t )
LC
 RC
 vC (t )  VS
2
dt
dt
d 2 vC (t ) R dvC (t ) 1
VS


vC (t ) 
2
dt
L dt
LC
LC
vC (t  to )  vt (t  to )  vss (t  to ) when t  to
Set of Solutions when t > to
 Similar to the solutions for the natural response, there
are three different solutions. To determine which one
to use, you need to calculate the natural angular
frequency of the series RLC network and the term a.
o 
1
LC
R
a
2L
Transient Solutions when t > to
s1Dt
s2 Dt
v
(
t
)

A
e

A
e
 Overdamped response (a > o) C
1
2
where t-to = Dt
s  a  a 2   2
1
0
s2  a  a 2  02
 Critically damped response (a = o)
vC (t )  ( A1  A2Dt )eaDt
 Underdamped response (a < o)
vC (t )  [ A1 cos(d Dt )  A2 sin(d Dt )]e aDt
 d  o 2  a 2
Steady State Solutions when t > to
 The final condition of the voltages across the capacitor
is the steady state solution.
 vC(∞s) = Vs
Complete Solution when t > to
 Overdamped response
vC (t )  A1e
s1Dt
 A2e
s2Dt
 Vs
 Critically damped response
vC (t )  ( A1  A2Dt )eaDt  Vs
 Underdamped response
vC (t )  [ A1 cos(d Dt )  A2 sin(d Dt )]eaDt  Vs
where Dt  t  to
Other Voltages and Currents
 Once the voltage across the capacitor is known, the
following equations for the case where t > to can be
used to find:
dvC (t )
iC (t )  C
dt
i (t )  iC (t )  iL (t )  iR (t )
diL (t )
vL (t )  L
dt
vR (t )  RiR (t )
Summary
 The set of solutions when t > to for the voltage across the
capacitor in a RLC network in series was obtained.
 The final condition for the voltage across the capacitor is the
steady state solution.
 Selection of equations is determine by comparing the natural
frequency o to a.
 Coefficients are found by evaluating the equation and its first
derivation at t = to- and t = ∞s.
 The voltage across the capacitor is equal to the initial
condition when t < to
 Using the relationships between current and voltage, the
current through the capacitor and the voltages and
currents for the inductor and resistor can be calculated.
Parallel RLC Network
Objective of Lecture
 Derive the equations that relate the voltages across a
resistor, an inductor, and a capacitor in parallel as:
 the unit step function associated with voltage or current
source changes from 0 to 1 or
 a switch connects a voltage or current source into the
circuit.
 Describe the solution to the 2nd order equations when
the condition is:
 Overdamped
 Critically Damped
 Underdamped
Parallel RLC Network
 With a current source switched into 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.
 Since the voltage source has a magnitude of 0V at t < to
 iL(to-) = 0A and v(to-) = vC(to-) = 0V
 vL(to-) = 0V and iC(to-) = 0A
 Once the steady state is reached after the voltage source
has a magnitude of Vs at t > to, replace the capacitor with
an open circuit and the inductor with a short circuit.


iL(∞s) = Is and v(∞s) = vC(∞s) = 0V
vL(∞s) = 0V and iC(∞s) = 0A
Selection of Parameter
 Initial Conditions
 iL(to-) = 0A and v(to-) = vC(to-) = 0V
 vL(to-) = 0V and iC(to-) = 0A
 Final Conditions
 iL(∞s) = Is 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).
Kirchhoff’s Current Law
iR (t )  iL (t )  iC (t )  iS (t )
v(t )  vR (t )  vL (t )  vC (t )
dvC (t )
vR (t )
 iL (t )  C
 IS
R
dt
diL (t )
vL (t )  v(t )  L
dt
d 2iL (t ) L diL (t )
LC

 iL (t )  I S
2
dt
R dt
d 2iL (t ) 1 diL (t ) iL (t ) I S



2
dt
RC dt
LC LC
iL (t )  it (t )  iss (t )
Set of Solutions when t > to
 Similar to the solutions for the natural response, there
are three different solutions. To determine which one
to use, you need to calculate the natural angular
frequency of the parallel RLC network and the term a.
1
o 
LC
1
a
2 RC
Transient Solutions when t > to
 Overdamped response
iL (t )  A1e
s1Dt
 A2e
s2Dt
 Critically damped response
 Underdamped response
iL (t )  ( A1  A2Dt )e
aDt
iL (t )  [ A1 cos(d Dt )  A2 sin(d Dt )]e
where
Dt  t  to
aDt
Other Voltages and Currents
 Once the 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
Complete Solution when t > to
 Overdamped response
iL (t )  A1e
s1Dt
s2 Dt
 Is
aDt
 Is
 A2e
 Critically damped response
iL (t )  ( A1  A2Dt )e
 Underdamped response
iL (t )  [ A1 cos(d Dt )  A2 sin(d Dt )]eaDt  Is
Summary
 The set of solutions when t > to for the current through the
inductor in a RLC network in parallel was obtained.
 The final condition for the current through the inductor is the
steady state solution.
 Selection of equations is determine by comparing the natural
frequency o to a.
 Coefficients are found by evaluating the equation and its first
derivation at t = to- and t = ∞s.
 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.