SME
University Physics II: Electricity & Magnetism
AC circuits
Use this for “A series circuit of 40 Ω, 0.1 H and 20 μF runs on 230 V at 50 Hz. Find each reactance, the impedance, the current and the phase angle.”
Diagram
X = XL − XC = 75.4 − 53.05
X = 22.35 Ω
Z
=
√
30
2
+
22.35
2
=
37.41
Ω.
φ = 36.68°: Z = 37.41∠36.68° Ω.
The current lags the voltage by 36.68°.
This is an example. Type your own values to start a new problem.
R
Resistance
example
L
Inductance
example
C
Capacitance
example
f
Frequency
example
XL
Inductive reactance
calculated
XC
Capacitive reactance
calculated
X
Net reactance
worked out (not typed)
Z
Impedance
calculated
V
Rms voltage
example
I
Rms current
calculated
φ
Phase angle of V ahead of I
calculated · °
Formulas
XL
=
2π
×
f
×
L
75.3982 = 2π × 60 × 0.2
XC
=
1
2π ×
f
×
C
×
10
-6
53.0516
=
1
2π × 60 × 50 ×
10
-6
X
=
XL
−
XC
+22.3466 = 75.3982 − 53.0516
Z
=
√
R
2
+
X
2
37.4081
=
√
30
2
+
22.3466
2
I
=
V
Z
3.2079
=
120
37.4081
φ
=
tan⁻¹(
X
R
)
+36.6819
=
tan⁻¹(
22.3466
30
)
Assumptions
•
V and I are rms values (Vᵣₘₛ = V₀ ÷ √2), the ones a meter reads; the same current flows through all three parts.
•
The inductor’s voltage leads the current by 90° and the capacitor’s lags it by 90°, so their reactances subtract: X = XL − XC.
•
φ is the angle of V ahead of I: φ > 0 means the current lags (XL wins), φ < 0 that it leads (XC wins).
•
C is typed in μF, so the lines write it × 10⁻⁶ to give farads.
Step-by-step
WE KNOW
R = 30 Ω, L = 0.2 H, C = 50 μF, f = 60 Hz, V = 120 V
FIND
inductive reactance (XL), capacitive reactance (XC), net reactance (X), impedance (Z), rms current (I), phase angle of V ahead of I (φ)
Step 1 · Find inductive reactance (XL)
Use XL = 2π × f × L
The
inductor
fights
a
change
in
current:
the
faster
the
swing
(2πf)
and
the
bigger
L
,
the
more
it
opposes.
XL = 2π × 60 × 0.2
XL = 75.3982 Ω
Step 2 · Find capacitive reactance (XC)
Use XC = 1 ÷ (2π × f × C × 10⁻⁶)
The
capacitor
passes
a
fast
swing
easily:
its
reactance
falls
as
f
or
C
grows.
μF
×
10
-6
gives
farads.
XC
=
1
2π × 60 × 50 ×
10
-6
XC = 53.0516 Ω
Step 3 · Find net reactance (X)
Use X = XL − XC
The two reactances point opposite ways on the phasor diagram, so they subtract.
X
=
75.3982
−
53.0516
X
=
+22.3466
Ω
Step 4 · Find impedance (Z)
Use Z = √(R² + X²)
R
and
X
are
at
right
angles
on
the
phasor
diagram,
so
Pythagoras
gives
the
size
of
Z
.
Z
=
√
30
2
+
22.34658
2
Z
=
√
900 + 499.4
Z
=
√
1
,399.4
Z
=
37.4081
Ω
Step 5 · Find rms current (I)
Use I = V ÷ Z
Ohm’s law for AC: the impedance takes the place of the resistance.
I
=
120
37.4081
I
=
3.2079
A
Step 6 · Find phase angle of V ahead of I (φ)
Use φ = tan⁻¹(X ÷ R)
Z
’s
angle
from
the
R
axis:
its
rise
X
over
its
run
R
.
φ
=
tan⁻¹(
22.3466
30
)
φ
=
+36.6819°
Check
Put the numbers back in. Both sides should match.
75.3982 = 2π × 60 × 0.2 ✓
53.0516
=
1
2π × 60 × 50 ×
10
-6
✓
+22.3466 = 75.3982 − 53.0516 ✓
37.4081
=
√
30
2
+
22.34658
2
✓
3.2079
=
120
37.4081
✓
+36.6819
=
tan⁻¹(
22.3466
30
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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