SME
Circuit Analysis II
AC power
Use this for “A load draws 10 A at 120 V with power factor 0.8 lagging. Find P, Q and S.”
Diagram
S
2
=
P
2
+
Q
2
:
S
=
√
960
2
+
720
2
=
1,200
VA.
The
angle
at
the
origin,
θ
=
+36.87°:
cos
θ
=
P
S
=
0.8
=
pf.
Drawn to scale: Q up from the end of P, S closing the triangle.
This is an example. Type your own values to start a new problem.
S (VA)
=
Vᵣₘₛ (V)
×
Iᵣₘₛ (A)
P (W)
=
S (VA)
×
pf
Q (var)
=
S (VA)
×
sin(
θ
°)
Formulas
S
=
Vᵣₘₛ
×
Iᵣₘₛ
1,200 = 120 × 10
pf
=
cos(
θ
)
0.8 = cos(36.8699°)
P
=
S
×
pf
960 = 1,200 × 0.8
Q
=
S
×
sin(
θ
)
+720 = 1,200 × sin(36.8699°)
S
2
=
P
2
+
Q
2
1,
200
2
=
960
2
+
720
2
Assumptions
•
Vᵣₘₛ and Iᵣₘₛ are rms values, the numbers an AC meter reads.
•
The angle θ is the voltage’s angle minus the current’s angle, between −90° and 90°.
•
Q > 0 is an inductive (lagging) load; Q < 0 is a capacitive (leading) load.
•
S, P and Q form a right triangle: P along, Q up and S the hypotenuse, so S² = P² + Q².
Step-by-step
WE KNOW
Vᵣₘₛ = 120 V, Iᵣₘₛ = 10 A, pf = 0.8
FIND
apparent power (S), power angle (θ), real power (P), reactive power (Q)
Step 1 · Find apparent power (S)
Use S = Vᵣₘₛ × Iᵣₘₛ
The apparent power is the rms voltage times the rms current, ignoring the angle.
S
=
120
×
10
S
=
1,200
VA
Step 2 · Find power angle (θ)
Use pf = cos(θ)
Two
angles
share
this
cosine,
±
θ
:
the
positive
one
is
a
lagging
load
(type
a
negative
θ
for
leading).
θ
=
cos⁻¹(pf)
θ
=
cos⁻¹(0.8)
θ
=
+36.8699°
Step 3 · Find real power (P)
Use P = S × pf
Only
the
part
of
S
in
step
with
the
voltage
does
work:
P
=
S
cos
θ
.
P
=
1,200
×
0.8
P
=
960
W
Step 4 · Find reactive power (Q)
Use Q = S × sin(θ)
The
leg
of
the
triangle
at
right
angles
to
P
:
the
power
that
sloshes
back
and
forth.
Q
=
1,200
×
sin(36.8699°)
Q
=
1,200
×
0.6
Q
=
+720
var
Check
Put the numbers back in. Both sides should match.
1,200 = 120 × 10 ✓
0.8 = cos(36.8699°) ✓
960 = 1,200 × 0.8 ✓
+720 = 1,200 × sin(36.8699°) ✓
1,
200
2
=
960
2
+
720
2
✓
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Parallel RC: impedance from admittance
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The power a load impedance draws
Topic check
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New questions each time. Pass to mark it complete.
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