RLC Exam like Example
Consider this circuit where the switch has been closed long enough
for the circuit to stabilized. Then open switch at t=0.
(a) Write the DE for circuit & initial conditions (current or voltage
across each element).
(b) Solve DE for I through R1
(c) What type of damping does this have? What R1 is needed for
the other damping types?
(d) For the (c) which is underdamped what is the 3 natural freq.,
damped freq. and damping factor?
(a) Using KVL
L
di
1
+ iR + ∫ i ⋅ dt = 0
C
dt
• Differentiating
di 1
d 2i
L 2 +R + =0
dt C
dt
RLC Example Cont’d
• Initial conditions, C is fully charged, acts a V source
C has VC = 40V
• L acts as a short before switch opens (no change in I)
• Therefore, current controlled by R
IL = IR =
V 40
=
= 2A
R 20
VL = 0 , VR = IR = 40V
(b) Assuming an exponential solution then get the equations
I (t ) = Ae st
0 = s 2 L + sR +
1
C
RLC Example Cont’d
• Solving this quadratic
0 = s2 + s
R 1
+
L CL
• The roots are
1
2
R ⎡⎛ R ⎞
1 ⎤
± ⎢⎜ ⎟ −
⎥
2 L ⎣⎝ 2 L ⎠ LC ⎦
R
20
=
= 2.5 × 10 3 sec −1
−3
2 L 2 × 4 × 10
1
1
= 2.5 × 107 sec − 2
=
−3
−5
LC 4 × 10 × 10
2
s=
• Thus the discriminate
⎡⎛ R ⎞ 2 1 ⎤
3 2
(
)
=
2
.
5
×
10
− 2.5 × 107 = −1.875 × 107 sec − 2
−
⎢⎜ ⎟
⎥
⎣⎝ 2 L ⎠ LC ⎦
[
• Because the discriminate is negative.
• This is an underdamped solution
]
RLC Example Cont’d
• Underdamped solved using ‘α, ω’ form
α=
ω n2 =
R
= 2.5 × 10 3 sec −1
2L
1
= 2.5 × 10 7 = 5000 rad / sec
LC
∴ω 2 = ω n2 − α 2 = 2.5 × 107 − 6.25 × 10 6 = 1.875 × 107
ω = 4.33 × 10 3
• Thus the solution will be of the form
[
i (t ) = e −αt A1e jωt + A2 e − jωt
• At t = 0, I = 2A
[
]
]
i (0 ) = e 0 A1e 0 + A2 e 0 = A1 + A2
A1 + A2 = 2 A
RLC Example Cont’d
• As 2nd order also need the derivative at t = 0
L
L
di
+ I 0 R = VC
dt
di
= VC − I 0 R = 0
dt
di
=0
dt
∴
di (0 )
= −αe −αt A1e jωt + A2 e − jωt + eαt jωA1e jωt − jωe − jωt
dt
[
]
[
= −α [ A1 + A2 ] + jω [ A1 − A2 ] = 0
∴ jω [ A1 − A2 ] = −α [ A1 + A2 ]
−α
2.5 × 10 3
[ A1 + A2 ] = j
[2]
A1 − A2 =
3
jω
4.33 × 10
• Thus
A1 + A2 = 2
A1 − A2 = j1.155
]
RLC Example Cont’d
• Adding
2 A1 = 2 + j1.155
A1 = 1 + j0.577
• Subtracting
2 A2 = 2 − j1.155
A2 = 1 − j0.577
∴ i(t ) = e
−αt
⎧⎪ jωt
⎫
− jωt
− jωt ⎪
jωt
+ e43 + 0.577 j e − e
e 42
⎨1
14
4244
3 ⎬⎪
⎪⎩ 2 cosωt
⎭
2 sin ωt
[
]
[
i(t ) = e − 2500 t [2 cos ωt + 1.155 sin ωt ]
• Alternate solution - assume
i (t ) = Ae −αt cos (ωt + θ )
l
• At t = 0
A cosθ = 2
A=
2
cos θ
]
RLC Example Cont’d
di
= −αe0 cos θ − ωe0 sin θ
dt
= −α cosθ − ω sinθ = 0
cos θ = −
ω
sin θ
α
α
sinθ
= tanθ = −
ω
cos θ
θ = tan −1 (− 0.577 ) = −30°
A=
2
= 2.31
cos(− 30°)
∴ i (t ) = 2.31e − 2500 t cos (ωt − 30° )
(c) This was underdamped, for critical damping
2
1
⎛ R ⎞
⎜ ⎟ =
LC
⎝ 2L ⎠
thus
⎛ (2 L )
R = ⎜⎜
⎝ LC
2
1
2
1
2
⎞
⎛ 8 × 10
4L
⎟ = ⎛⎜ ⎞⎟ = ⎜⎜
−5
⎟
⎝C ⎠
⎝ 10
⎠
R = 40Ω
−3
⎞
⎟⎟
⎠
1
2
• For overdamped needs R1 >40Ω - say 60Ω
• Thus
R
α=
= 7.5 × 10 3 sec −1
2L
• Thus the discriminate
⎡⎛ R ⎞ 2 1 ⎤
3 2
(
)
−
=
7
.
5
×
10
− 2.5 × 107 = 3.125 × 107
⎜
⎟
⎢
⎥
⎣⎝ 2 L ⎠ LC ⎦
• Thus the solution is
s1 = α + 3.125 x107 = 13090 sec −1
[
]
s2 = α − 3.125 x107 = 1909 sec −1
i (t ) = A1e −13090 t + A2 e −1909 t
[
]