SME
University Physics II: Electricity & Magnetism · Problem type
Average power and the power factor
Use this for “A coil of 40 Ω resistance and 30 Ω reactance runs on 230 V rms. Find the current, the power factor and the average power it takes.”
Diagram
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.
I (
)
=
V (
)
(
R (
)
)
2
+
(
X (
)
)
2
Z
Impedance
calculated
φ
Phase angle
calculated · °
cos φ
Power factor
calculated
P
Average power
calculated
Formulas
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
)
cos
φ
=
R
Z
0.802
=
30
37.4081
P
=
V
×
I
×
(cos
φ
)
308.7105 = 120 × 3.2079 × 0.802
P
=
I
2
×
R
308.7105
=
3.2079
2
×
30
Assumptions
•
V and I are rms values (Vᵣₘₛ = V₀ ÷ √2), and one current runs through every part of the series circuit.
•
X is the net reactance XL − XC: positive when the inductor wins (the current lags), negative when the capacitor wins (it leads).
•
Only the resistance takes power on average; the reactance stores energy and gives it back each cycle.
•
So the average power is P = VI cos φ = I²R. The power factor cos φ = R ÷ Z is between 0 and 1, the same for φ and −φ.
Step-by-step
WE KNOW
V = 120 V, R = 30 Ω, X = +22.3466 Ω
FIND
impedance (Z), rms current (I), phase angle (φ), power factor (cos φ), average power (P)
Step 1 · Find impedance (Z)
Use Z = √(R² + X²)
R
and
X
are
at
right
angles
on
the
complex
plane:
Pythagoras
gives
the
impedance.
Z
=
√
30
2
+
22.3466
2
Z
=
√
900 + 499.4
Z
=
√
1
,399.4
Z
=
37.408147
Ω
Step 2 · Find rms current (I)
Use I = V ÷ Z
Ohm’s law for AC: the rms voltage over the impedance.
I
=
120
37.408147
I
=
3.2078574
A
Step 3 · Find phase angle (φ)
Use φ = tan⁻¹(X ÷ R)
The
angle
of
Z
=
R
+
jX
from
the
real
axis:
the
reactance
over
the
resistance.
φ
=
tan⁻¹(
22.3466
30
)
φ
=
+36.6819°
Step 4 · Find power factor (cos φ)
Use cos φ = R ÷ Z
The
power
factor
is
the
real
part
of
Z
over
its
size:
cos
φ
=
R
Z
.
cos
φ
=
30
37.408147
cos
φ
=
0.8019643
Step 5 · Find average power (P)
Use P = V × I × (cos φ)
Only
the
part
of
the
current
in
step
with
the
voltage,
I
cos
φ
,
carries
power
on
average.
P
=
120
×
3.2078574
×
0.8019643
P
=
384.9
×
0.802
P
=
308.7105
W
Check
Put the numbers back in. Both sides should match.
37.408147
=
√
30
2
+
22.3466
2
✓
3.2078574
=
120
37.408147
✓
+36.6819
=
tan⁻¹(
22.3466
30
)
✓
0.8019643
=
30
37.408147
✓
308.7105 = 120 × 3.2078574 × 0.8019643 ✓
308.7105
=
3.2078574
2
×
30
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.