SME
Circuit Analysis II · Problem type
Reactance and resonance at a frequency
Use this for “Find the reactances of a 0.1 H inductor and a 100 μF capacitor at 60 Hz, and the frequency at which they are equal.”
Diagram
X = XL − XC = 37.7 − 26.53
X = 11.17 Ω
|jX|
=
√
0
2
+
11.17
2
=
11.17
Ω.
θ = 90°: jX = 11.17∠90° Ω.
The current lags the voltage by 90°.
This is an example. Type your own values to start a new problem.
XL (Ω)
=
2π
×
f (Hz)
×
L (H)
XC (Ω)
=
1
2π
×
f (Hz)
×
C (μF)
f₀ (Hz)
=
1
2π
L (H)
×
C (μF)
X
Net reactance in series
worked out (not typed)
Formulas
X
=
XL
−
XC
+11.1733 = 37.6991 − 26.5258
XL
=
2π
×
f
×
L
37.6991 = 2π × 60 × 0.1
XC
=
1
2π ×
f
×
C
26.5258
=
1
2π × 60 × 0.0001
f₀
=
1
2π ×
√
L
×
C
50.3292
=
1
2π ×
√
0.1 × 0.0001
Formulas are worked in F. The step-by-step shows the conversions.
Assumptions
•
The frequency f is in hertz; the angular frequency is ω = 2πf, so XL = 2πfL and XC = 1 ÷ (2πfC).
•
The inductor and the capacitor are ideal: no resistance in their wires or plates.
•
At the resonant frequency f₀ the two reactances are equal and cancel in series: XL = XC there.
Step-by-step
WE KNOW
f = 60 Hz, L = 0.1 H, C = 100 μF
FIND
inductive reactance (XL), capacitive reactance (XC), resonant frequency (f₀), net reactance in series (X)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to F).
C = 100 μF = 0.0001 F (1 F = 10⁶ μF)
Step 2 · Find inductive reactance (XL)
Use XL = 2π × f × L
The inductor’s reactance is ωL with ω = 2πf: it grows in step with the frequency.
XL = 2π × 60 × 0.1
XL = 37.6991 Ω
Step 3 · Find capacitive reactance (XC)
Use XC = 1 ÷ (2π × f × C)
The
capacitor’s
reactance
is
1
ωC
,
with
C
in
farads:
doubling
the
frequency
halves
it.
XC
=
1
2π × 60 × 0.0001
XC = 26.5258 Ω
Step 4 · Find resonant frequency (f₀)
Use f₀ = 1 ÷ (2π × √(L × C))
Set
2πfL
=
1
2πfC
and
solve
for
f
:
f
2
=
1
4π²
LC
,
with
C
in
farads.
f₀
=
1
2π ×
√
0.1 × 0.0001
f₀
=
50.3292
Hz
Step 5 · Find net reactance in series (X)
Use X = XL − XC
In
series
jXL
points
up
and
−jXC
down:
above
f₀
the
inductor
wins,
below
it
the
capacitor.
X
=
37.6991
−
26.5258
X
=
+11.1733
Ω
Check
Put the numbers back in. Both sides should match.
+11.1733 = 37.6991 − 26.5258 ✓
37.6991 = 2π × 60 × 0.1 ✓
26.5258
=
1
2π × 60 × 0.0001
✓
50.3292
=
1
2π ×
√
0.1 × 0.0001
✓
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