SME
Circuit Analysis I · Problem type
Damping of a series RLC circuit
Use this for “A series RLC has 10 Ω, 1 mH and 10 μF. Is it over-, under- or critically damped?”
Diagram
The switch closes at t = 0; R, L and C share one current round the loop.
This is an example. Type your own values to start a new problem.
α (s⁻¹)
=
R (Ω)
2
×
L (mH)
ω₀ (rad/s)
=
1
L (mH)
×
C (μF)
D
Damping
worked out (not typed)
ωd
Damped frequency
worked out (not typed) · rad/s
Formulas
α
=
R
2 ×
L
5,000
=
10
2 × 0.001
ω₀
=
1
÷
√
L
×
C
10,000
=
1
÷
√(0.001
×
(1
×
10
-5
))
D
:
compare
α
with
ω₀
5,000 < 10,000, so underdamped
ωd
=
√
ω₀
2
−
α
2
8,660.254
=
√
10,
000
2
− 5,
000
2
Formulas are worked in H, F. The step-by-step shows the conversions.
Assumptions
•
A series RLC is switched onto a DC source Vₛ at t = 0, and vC heads for Vₛ.
•
α = R ÷ (2L) and ω₀ = 1 ÷ √(LC): α > ω₀ is overdamped, α = ω₀ critically damped, α < ω₀ underdamped.
•
Underdamped, it rings at the damped frequency ωd = √(ω₀² − α²) inside a dying envelope.
Step-by-step
WE KNOW
R = 10 Ω, L = 1 mH, C = 10 μF
FIND
neper frequency (α), resonant frequency (ω₀), damping (D), damped frequency (ωd)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to H, F).
L = 1 mH = 0.001 H (1 H = 1000 mH)
C = 10 μF = 1 × 10⁻⁵ F (1 F = 10⁶ μF)
Step 2 · Find neper frequency (α)
Use α = R ÷ (2 × L)
The
neper
frequency:
how
fast
R
drains
the
energy
that
L
and
C
pass
back
and
forth.
α
=
10
2 × 0.001
α
=
10
0.002
α
=
5,000
s⁻¹
Step 3 · Find resonant frequency (ω₀)
Use ω₀ = 1 ÷ √(L × C)
The
frequency
at
which
L
and
C
trade
energy
with
no
R
:
henries
×
farads
gives
s
2
.
ω₀
=
1
÷
√
0.001 ×
10
-5
ω₀
=
1
÷
√
10
⁻⁸
ω₀
=
1
0.0001
ω₀
=
10,000
rad/s
Step 4 · Find damping (D)
Use D: compare α with ω₀
α
is
smaller
than
ω₀
:
complex
roots,
so
vC
overshoots
Vₛ
and
rings
as
it
settles.
5,000 < 10,000: underdamped
D
=
compare
5,000
with
10,000
D
=
underdamped
Step 5 · Find damped frequency (ωd)
Use ωd = √(ω₀² − α²)
Underdamped
only:
the
roots
are
−
α
±
jωd,
so
it
rings
a
little
slower
than
ω₀
.
ωd
=
√
10,
000
2
− 5,
000
2
ωd
=
√
100,000,000 − 25,000,000
ωd
=
√
75
,000,000
ωd = 8,660.254 rad/s
Check
Put the numbers back in. Both sides should match.
5,000
=
10
2 × 0.001
✓
10,000
=
1
÷
√
0.001 ×
10
-5
✓
5,000 < 10,000, so underdamped ✓
8,660.254
=
√
10,
000
2
− 5,
000
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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