SME
Circuit Analysis II · Problem type
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.”
Diagram
Iab = 12∠0° A, Ibc = 12∠−120° A, Ica = 12∠120° A, Iₐ = 20.78∠−30° A.
−Ica = 12∠−60° A, tip to tail: Iₐ = Iab − Ica.
This is an example. Type your own values to start a new problem.
Iₚₕ (A)
=
VL (V)
|Z| (Ω)
IL (A)
=
3
×
Iₚₕ (A)
P (kW)
=
3
×
VL (V)
×
Iₚₕ (A)
×
pf
Vₚₕ
Phase voltage
calculated
Formulas
Vₚₕ = VL
240 = 240
Iₚₕ = Vₚₕ ÷ |Z|
12
=
240
20
IL
=
√
3
×
Iₚₕ
20.7846
=
√
3
×
12
P
=
3
×
Vₚₕ
×
Iₚₕ
×
pf
8,640 = 3 × 240 × 12 × 1
Formulas are worked in W. The step-by-step shows the conversions.
Assumptions
•
In a Δ each phase impedance sits across two lines, so the phase voltage is the line voltage; values are rms.
•
Iₐ = Iab − Ica: √3 times a phase current and 30° behind Iab (drawn tip to tail, with j² = −1).
•
Balanced: each phase current trails its phase voltage by θ = cos⁻¹(pf), the same in all three.
Step-by-step
WE KNOW
VL = 240 V, |Z| = 20 Ω, pf = 1
FIND
phase voltage (Vₚₕ), phase current (Iₚₕ), line current (IL), real power (P)
Step 1 · Find phase voltage (Vₚₕ)
Use Vₚₕ = VL
Each phase sits across two lines.
Vₚₕ = 240 V
Step 2 · Find phase current (Iₚₕ)
Use Iₚₕ = Vₚₕ ÷ |Z|
Ohm’s law in one phase.
Iₚₕ
=
240
20
Iₚₕ = 12 A
Step 3 · Find line current (IL)
Use IL = √3 × Iₚₕ
Each
line
carries
the
difference
of
two
phase
currents
120°
apart:
√
3
times
one.
IL
=
√
3
×
12
IL = 1.732 × 12
IL = 20.7846 A
Step 4 · Find real power (P)
Use P = 3 × Vₚₕ × Iₚₕ × pf
Three phases, each taking Vₚₕ Iₚₕ × pf of real power.
P
=
3
×
240
×
12
×
1
P
=
720
×
12
P
=
8,640
W
Step 5 · Convert the answers
Change each answer back to the unit you chose.
P = 8,640 W = 8.64 kW (1 kW = 1000 W)
Check
Put the numbers back in. Both sides should match.
240 = 240 ✓
12
=
240
20
✓
20.7846
=
√
3
×
12
✓
8,640 = 3 × 240 × 12 × 1 ✓
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
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.”