Why is I_R + I_Y + I_B = 0?

The line currents are equal in magnitude and 120° apart, and KCL holds with no neutral path

2070 Chaitra · 8 marks

Delta load (15+j20) phase and line currents

Short answer

Phase voltage equals line voltage 220 V. Line currents: 15.24 A at −83.13°, 156.87°, 36.87°. Power per phase is 1161.6 W. Sum of three line currents is zero (balanced system with no neutral).

A 220 V, 3-phase voltage is applied to a balanced delta connected 3-phase load of phase impedance (15+j20)Ω. Calculate: i) The phase voltages ii) The phasor current in each line iii) The power consumed per phase iv) Draw the phasor diagram v) What is the phasor sum of three line currents? Why does it have this value?

Balanced delta load Z_ph = 15 + j20 = 25∠53.13° Ω, V_L = 220 V, RYB sequence assumed.

i) Phase voltages

In delta, the phase voltage equals the line voltage: V_ph = 220 V. V_RY = 220∠0°, V_YB = 220∠−120°, V_BR = 220∠120° V.

ii) Currents

Phase current: I_ph = 220/25 = 8.8 A, lagging its phase voltage by 53.13°.

I_RY = 8.8∠−53.13° A, I_YB = 8.8∠−173.13° A, I_BR = 8.8∠66.87° A

Line current magnitude: I_L = √3 × 8.8 = 15.24 A. In delta, each line current lags the corresponding phase current by 30°.

I_R = I_RY − I_BR = 15.24∠−83.13° A
I_Y = I_YB − I_RY = 15.24∠−203.13° = 15.24∠156.87° A
I_B = I_BR − I_YB = 15.24∠+36.87° A

iii) Power per phase

P_ph = I_ph² R = 8.8² × 15 = 1161.6 W (total 3484.8 W)

iv) Phasor diagram (V_RY as reference)

Phasor	Angle
V_RY, V_YB, V_BR	0°, −120°, +120°
I_RY, I_YB, I_BR (53.13° behind their voltages)	−53.13°, −173.13°, +66.87°
I_R, I_Y, I_B (30° behind the phase currents)	−83.13°, +156.87°, +36.87°

           V_BR (120°)      I_BR (66.9°)
               ↖            ↗
                 ↖        ↗      ↗ I_B (36.9°)
                   ↖    ↗    ↗
   I_Y (156.9°) ←──────  ●  ──────────────► V_RY (0°)
   I_YB (−173.1°)      ↙ │ ↘
                     ↙   │   ↘ I_RY (−53.1°)
                   ↙     ↓
         V_YB (−120°)    I_R (−83.1°)

The three line voltages are 120° apart, each phase current is 53.1° behind its voltage, and each line current is 30° behind its phase current.

v) Phasor sum of the line currents

I_R + I_Y + I_B = 0.

For a balanced load the three line currents have equal magnitude and are 120° apart, so they form a closed triangle. This is also required by Kirchhoff's current law, because a delta-connected three-wire system has no path (neutral) for any residual current.

✓ Answer checked by Claude Opus 5.5 (high effort)

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