SME
University Physics II: Electricity & Magnetism · Problem type
Resonance in a series RLC circuit
Use this for “A radio tuner has a 0.5 mH coil, a 200 pF capacitor and 10 Ω of resistance. At what frequency does it resonate, and how wide is its band?”
Chart
Level
1
÷
√
2
=
0.7071
meets
the
curve
at
u₁
=
0.7906,
u₂
=
1.265
This is an example. Type your own values to start a new problem.
f₀ (
)
=
1
2π
L (
)
×
C (
)
R
Resistance
example
Q
Quality factor
calculated
Δf
Bandwidth
calculated
f₁
Lower band edge
worked out (not typed)
f₂
Upper band edge
worked out (not typed)
Formulas
f₀
=
1
2π ×
√
L
×
C
×
10
-9
50.3292
=
1
2π ×
√
200 × 50 ×
10
-9
Q
=
(
1
R
)
×
√(
L
C
×
10
3
)
2.1082
=
(
1
30
)
×
√(
200
50
×
10
3
)
Δf
=
f₀
Q
23.8732
=
50.3292
2.1082
f₂
=
√(
f₀
2
+
(
Δf
2
)
2
)
+
Δf
2
63.662
=
√(
50.3292
2
+
(
23.8732
2
)
2
)
+
23.8732
2
f₁
=
f₀
2
f₂
39.7887
=
50.3292
2
63.662
Assumptions
•
One series loop: R, L and C carry the same current, and the source keeps its voltage while its frequency changes.
•
At f₀ the reactances cancel, 2πf₀L = 1 ÷ (2πf₀C), so Z = R and the current peaks at I₀ = V ÷ R.
•
The band edges f₁ and f₂ are where I = I₀ ÷ √2, so the power is half its peak; Δf = f₂ − f₁ = R ÷ (2πL), and f₁ × f₂ = f₀².
•
mH × μF × 10⁻⁹ gives s², and mH ÷ μF × 10³ gives Ω².
Step-by-step
WE KNOW
L = 200 mH, C = 50 μF, R = 30 Ω
FIND
resonant frequency (f₀), quality factor (Q), bandwidth (Δf), upper band edge (f₂), lower band edge (f₁)
Step 1 · Find resonant frequency (f₀)
Use f₀ = 1 ÷ (2π × √(L × C × 10⁻⁹))
Set
the
reactances
equal,
2πfL
=
1
2πfC
,
and
solve
for
f;
mH
×
μF
×
10
-9
gives
s
2
.
f₀
=
1
2π ×
√
200 × 50 ×
10
-9
f₀
=
50.32921
Hz
Step 2 · Find quality factor (Q)
Use Q = (1 ÷ R) × √(L ÷ C × 10³)
The
quality
factor
is
√(
L
C
)
over
R
:
mH
μF
×
10
3
gives
Ω
2
,
so
the
root
is
in
ohms
and
Q
has
no
unit.
Q
=
(
1
30
)
×
√(
200
50
×
10
3
)
Q
=
1
30
×
√(
200
50
×
1,000)
Q
=
1
30
×
√
4 × 1,000
Q
=
1
30
×
√
4
,000
Q
=
1
30
×
63.25
Q
=
0.03333
×
63.25
Q
=
2.10819
Step 3 · Find bandwidth (Δf)
Use Δf = f₀ ÷ Q
The
half-power
band
is
f₀
Q
wide:
the
higher
Q
,
the
narrower
the
band.
Δf
=
50.32921
2.10819
Δf = 23.8732 Hz
Step 4 · Find upper band edge (f₂)
Use f₂ = √(f₀² + (Δf ÷ 2)²) + Δf ÷ 2
The
edges
are
Δf
apart
with
f₁
×
f₂
=
f₀
2
,
so
f₂
is
half
the
band
above
√(
f₀
2
+
(
Δf
2
)
2
).
f₂
=
√(
50.32921
2
+
(
23.8732
2
)
2
)
+
23.8732
2
f₂
=
√
50.3292
2
+
11.9366
2
+
23.8732
2
f₂
=
√
2,533 + 142.5
+
23.8732
2
f₂
=
√
2
,675.5
+
23.8732
2
f₂
=
51.73
+
23.8732
2
f₂
=
51.73
+
11.9366
f₂
=
63.662
Hz
Step 5 · Find lower band edge (f₁)
Use f₁ = f₀² ÷ f₂
f₀
is
the
geometric
mean
of
the
edges,
so
f₁
=
f₀
2
f₂
(the
same
as
f₂
−
Δf).
f₁
=
50.32921
2
63.662
f₁
=
2,533
63.662
f₁
=
39.7887
Hz
Check
Put the numbers back in. Both sides should match.
50.32921
=
1
2π ×
√
200 × 50 ×
10
-9
✓
2.10819
=
(
1
30
)
×
√(
200
50
×
10
3
)
✓
23.8732
=
50.32921
2.10819
✓
63.662
=
√(
50.32921
2
+
(
23.8732
2
)
2
)
+
23.8732
2
✓
39.7887
=
50.32921
2
63.662
✓
Topic check
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