SME
Circuit Analysis II · Problem type
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.”
Diagram
ZY
=
ZΔ
3
=
30
Ω
3
=
10
Ω
IΔ
=
VL
ZΔ
=
208
V
30
Ω
=
6.933
A
IL
=
VL
√
3
× ZY
=
208
V
√
3
× 10 Ω
IL = 12.01 A
Balanced: each Y arm is a third of a Δ side, so the same line voltages draw the same line currents.
This is an example. Type your own values to start a new problem.
ZY (Ω)
=
ZΔ (Ω)
3
IL (A)
=
VL (V)
3
×
ZY (Ω)
IΔ
Delta phase current
calculated
Formulas
ZY
=
ZΔ
3
10
=
30
3
IΔ
=
VL
ZΔ
6.9333
=
208
30
IL
=
VL
√
3
× ZY
12.0089
=
208
√
3
× 10
IL
=
√
3
×
IΔ
12.0089
=
√
3
×
6.9333
Assumptions
•
Balanced: the three Δ impedances are equal, so each arm of the equivalent Y is ZY = ZΔ ÷ 3.
•
Equivalent means the same line voltages draw the same line currents; in the Y the line current is the phase current.
•
In the Δ, Iₐ = Iab − Ica is √3 times the phase current IΔ (drawn tip to tail, with j² = −1).
Step-by-step
WE KNOW
ZΔ = 30 Ω, VL = 208 V
FIND
wye impedance (ZY), delta phase current (IΔ), line current (IL)
Step 1 · Find wye impedance (ZY)
Use ZY = ZΔ ÷ 3
Match the impedance between two lines: 2ZY in the Y, and ZΔ in parallel with 2ZΔ, which is 2ZΔ ÷ 3, in the Δ.
ZY
=
30
3
ZY = 10 Ω
Step 2 · Find delta phase current (IΔ)
Use IΔ = VL ÷ ZΔ
Each Δ impedance has the full line voltage across it.
IΔ
=
208
30
IΔ = 6.9333 A
Step 3 · Find line current (IL)
Use IL = VL ÷ (√3 × ZY)
Each
Y
arm
has
VL
÷
√
3
across
it,
and
its
current
is
the
line
current.
IL
=
208
√
3
× 10
IL
=
208
1.732 × 10
IL
=
208
17.32
IL = 12.0089 A
Check
Put the numbers back in. Both sides should match.
10
=
30
3
✓
6.9333
=
208
30
✓
12.0089
=
208
√
3
× 10
✓
12.0089
=
√
3
×
6.93333
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Three-phase circuits
Main lesson
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.”
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.”