Hi, we all know Vline is equal to Vphase and ILine = √3.Iphase in delta connected induction motors.
And let's say (delta connected) Vs=400V and Zeq = 100 ohm
Is = Vs/Zeq = 400/100 = 4A
what's this Is current, Is = ILine or Is = Iphase ?
Question says find the current drawn by the motor from the network but i don't know if i take Is directly or multiply it by √3

 

Hi, we all know Vline is equal to Vphase and ILine = √3.Iphase in delta connected induction motors.
And let's say (delta connected) Vs=400V and Zeq = 100 ohm
Is = Vs/Zeq = 400/100 = 4A
what's this Is current, Is = ILine or Is = Iphase ?
Question says find the current drawn by the motor from the network but i don't know if i take Is directly or multiply it by √3

Hi,
The equations that you have quoted in your first line are quite correct.
For 3 Phase Delta connections: Vline is the same value as Vphase and Iline = 1.732 x Iphase.
First determine the current that will flow through a phase winding and then multiply this by 1.732
Iphase will be Vphase / Zphase, in the case of delta connection this will be the same as Vline/ Zphase
Your problem would seem to be, is what you are calling Zeq the Zphase value? If it is then multiply the 4A by 1,732 to obtain the line current value. This would give 1.732 x 4A = 6.928A drawn on each line of the 3 phase supply, (known as the supply current, Is). Some students think that the line current should be 2 x the phase current because the line current flows into two windings. However, the two phase currents that are derived from the line current are displaced in time by 120 electrical degrees. Hence 1.732 times instead of 2 times.
Hope this helps with your understanding of the subject
Colin