SME
Circuit Analysis II
Three-phase circuits
Use this for “A balanced Y load of 12 Ω per phase at pf 0.8 is fed at 208 V line to line. Find the line current and power.”
Diagram
Vₐₙ = 120.1∠0° V, Vbn = 120.1∠−120° V, Vcn = 120.1∠120° V, Iₐ = 10.01∠−36.87° A.
Vab = 208∠30° V, tip to tail: Vₐₙ = Vbn + Vab.
Phasors in A are drawn to their own scale.
θ = 36.87° between Vₐₙ and Iₐ.
This is an example. Type your own values to start a new problem.
IL (A)
=
VL (V)
3
×
|Z| (Ω)
S (kVA)
=
3
×
VL (V)
×
IL (A)
P (kW)
=
S (kVA)
×
pf
Vₚₕ
Phase voltage
calculated
θ
Power factor angle
calculated · °
Q
Reactive power
calculated
Formulas
Vₚₕ
=
VL
÷
√
3
120.0889
=
208
÷
√
3
IL = Vₚₕ ÷ |Z|
10.0074
=
120.0889
12
pf
=
cos(
θ
)
0.8 = cos(36.8699°)
S
=
√
3
×
VL
×
IL
3,605.3333
=
√
3
×
208
×
10.0074
P
=
S
×
pf
2,884.2667 = 3,605.3333 × 0.8
Q
=
S
×
sin(
θ
)
2,163.2 = 3,605.3333 × sin(36.8699°)
Formulas are worked in VA, W, var. The step-by-step shows the conversions.
Assumptions
•
Balanced source and load: three equal phase voltages 120° apart, three equal impedances; every value is rms, and j² = −1.
•
In a Y the line current is the phase current, and Vab = Vₐₙ − Vbn is √3 times a phase voltage, 30° ahead of Vₐₙ.
•
The load lags: each phase current trails its phase voltage by θ = cos⁻¹(pf).
Step-by-step
WE KNOW
VL = 208 V, |Z| = 12 Ω, pf = 0.8
FIND
phase voltage (Vₚₕ), line current (IL), power factor angle (θ), apparent power (S), real power (P), reactive power (Q)
Step 1 · Find phase voltage (Vₚₕ)
Use Vₚₕ = VL ÷ √3
Each
phase
sits
between
a
line
and
neutral;
the
line-to-line
voltage
is
√
3
times
it.
Vₚₕ
=
208
÷
√
3
Vₚₕ
=
208
1.732
Vₚₕ = 120.088856 V
Step 2 · Find line current (IL)
Use IL = Vₚₕ ÷ |Z|
Ohm’s law in one phase; in a Y that phase current is also the line current.
IL
=
120.088856
12
IL = 10.0074047 A
Step 3 · Find power factor angle (θ)
Use pf = cos(θ)
Undo the cosine; the current lags by this angle.
θ
=
cos⁻¹(pf)
θ
=
cos⁻¹(0.8)
θ
=
36.8699°
Step 4 · Find apparent power (S)
Use S = √3 × VL × IL
Three
phases
take
Vₚₕ
IL
each,
and
3Vₚₕ
=
√
3
VL.
S
=
√
3
×
208
×
10.0074047
S
=
1.732
×
208
×
10.0074
S
=
360.256
×
10.0074
S
=
3,605.3333
VA
Step 5 · Find real power (P)
Use P = S × pf
The
real
power
is
the
in-step
share
of
S
.
P
=
3,605.33333
×
0.8
P
=
2,884.2667
W
Step 6 · Find reactive power (Q)
Use Q = S × sin(θ)
The
reactive
power
is
the
share
of
S
that
is
90°
out
of
step.
Q
=
3,605.33333
×
sin(36.8699°)
Q
=
3,605.3333
×
0.6
Q
=
2,163.2
var
Step 7 · Convert the answers
Change each answer back to the unit you chose.
S = 3,605.3333 VA = 3.605333 kVA (1 kVA = 1000 VA)
P = 2,884.2667 W = 2.8843 kW (1 kW = 1000 W)
Q = 2,163.2 var = 2.1632 kvar (1 kvar = 1000 var)
Check
Put the numbers back in. Both sides should match.
120.088856
=
208
÷
√
3
✓
10.0074047
=
120.088856
12
✓
0.8 = cos(36.8699°) ✓
3,605.33333
=
√
3
×
208
×
10.0074047
✓
2,884.267 = 3,605.33333 × 0.8 ✓
2,163.2 = 3,605.33333 × sin(36.8699°) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Balanced delta load
Use this for “A balanced Δ load of 20 Ω per phase at pf 1 is fed at 240 V. Find the phase and line currents and the power.”
Delta to wye conversion
Use this for “Replace a balanced Δ load of 30 Ω per phase with its equivalent Y, and find the line current at 208 V.”
Two-wattmeter method
Use this for “Two wattmeters on a balanced three-phase load read 1000 W and 2000 W. Find the total power and the power factor.”