SME
Circuit Analysis II · Problem type
Power-factor correction with a capacitor
Use this for “A 960 W load runs at power factor 0.8 lagging on 120 V, 60 Hz. What capacitor raises the power factor to 0.95?”
Diagram
S₁ = ⟨960, 720⟩
|S₁| = 1,200.
QC: 404.5 at −90°, so QC = ⟨0, −404.5⟩.
Tip to tail: S₂ = ⟨960, 315.5⟩
|S₂| = 1,011 at 18.19°.
This is an example. Type your own values to start a new problem.
P
Real power
example
pf₁
Power factor before
example
pf₂
Power factor after
example
Vᵣₘₛ
RMS voltage
example
f
Frequency
example · Hz
Q₁
Reactive power before
calculated
Q₂
Reactive power after
worked out (not typed)
QC
Capacitor’s reactive power
calculated
C
Capacitance
calculated
Formulas
Q₁
=
P
×
tan(cos⁻¹(pf₁))
720 = 960 × tan(cos⁻¹(0.8))
Q₂
=
P
×
tan(cos⁻¹(pf₂))
315.5367 = 960 × tan(cos⁻¹(0.95))
QC
=
Q₁
−
Q₂
404.4633 = 720 − 315.5367
C
=
QC
2π ×
f
×
Vᵣₘₛ
2
7.4505 ×
10
-5
=
404.4633
2π × 60 ×
120
2
Formulas are worked in F. The step-by-step shows the conversions.
Assumptions
•
The capacitor is in parallel with the load, so the load’s P and the voltage Vᵣₘₛ do not change.
•
Both power factors are lagging (inductive load); correction never goes past 1, so pf₂ ≥ pf₁.
•
The capacitor supplies QC = V²ωC = 2πfV²C of reactive power, which the source no longer has to.
Step-by-step
WE KNOW
P = 960 W, pf₁ = 0.8, pf₂ = 0.95, Vᵣₘₛ = 120 V, f = 60 Hz
FIND
reactive power before (Q₁), reactive power after (Q₂), capacitor’s reactive power (QC), capacitance (C)
Step 1 · Find reactive power before (Q₁)
Use Q₁ = P × tan(cos⁻¹(pf₁))
θ₁
=
cos⁻¹
pf₁
is
the
load’s
angle;
in
the
power
triangle
Q
is
P
times
tan
θ.
Q₁
=
960
×
tan(cos⁻¹(0.8))
Q₁
=
720
var
Step 2 · Find reactive power after (Q₂)
Use Q₂ = P × tan(cos⁻¹(pf₂))
The
same
triangle
at
the
new
power
factor:
P
stays,
the
angle
shrinks.
Q₂
=
960
×
tan(cos⁻¹(0.95))
Q₂
=
315.5367
var
Step 3 · Find capacitor’s reactive power (QC)
Use QC = Q₁ − Q₂
The capacitor supplies the reactive power the source no longer has to.
QC = 720 − 315.5367
QC = 404.4633 var
Step 4 · Find capacitance (C)
Use C = QC ÷ (2π × f × Vᵣₘₛ²)
A
capacitor
on
Vᵣₘₛ
supplies
QC
=
V
2
ωC
with
ω
=
2πf,
so
divide
QC
by
2πfV².
C
=
404.4633
2π × 60 ×
120
2
C
=
7.4505 ×
10
-5
F
Step 5 · Convert the answers
Change each answer back to the unit you chose.
C = 7.4505 × 10⁻⁵ F = 74.505 μF (1 F = 10⁶ μF)
Check
Put the numbers back in. Both sides should match.
720 = 960 × tan(cos⁻¹(0.8)) ✓
315.5367 = 960 × tan(cos⁻¹(0.95)) ✓
404.4633 = 720 − 315.5367 ✓
7.4505 ×
10
-5
=
404.4633
2π × 60 ×
120
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
AC power
Main lesson
The power a load impedance draws
Use this for “A 120 V rms source feeds a load Z = 8 + j6 Ω. Find the current and the real and reactive power.”
RMS voltage from the peak
Use this for “A sine wave peaks at 170 V. Find its rms value and the average power it delivers to a 60 Ω resistor.”