Home › Electrical Engineering Forum › General Discussion › Inductive load currents at varying voltages
 This topic has 4 replies, 3 voices, and was last updated 9 years, 4 months ago by Spir Georges GHALI.

AuthorPosts

2012/06/04 at 11:01 am #10832sturnbull720Participant
Just wondering if anyone can help me, I have recently come across a large voltage drop issue when a client is using a mig welder. When the welder is in operation at full load current it draws 26 amps (single phase load). The voltage at the outlet at no load is 242V and under full load current is 208V. The voltage drop is due to a preexisting large run (over 50 metres) of 2.5mm2 twin and earth cable which is grossly under rated for its application. I am looking at upgrading the circuit cabling to larger CSA conductors which will overcome the voltage drop issue. My question is however, if the welder is drawing 26 amps at 208V, what will the current draw be at nominal voltage of approx 240V? I understand with a purely resistive load that, impedence remaining constant, as voltage decreases, current will do the same. I’m trying to avoid installing circuit wiring suitable for circuit length/voltage drop/load current, only to find out at nominal voltage the load will be considerably higher. Any information will help. Thanks
2012/06/04 at 2:01 pm #12962smithwalkerParticipantThe large voltage drop across a load is equal to the source voltage. Which means that current enters a lamp at 120 V and leaves that lamp at 0 V. Sometimes the wires themselves have enough resistance to be part of the load, so some of the voltage is dropped just overcoming that resistance. This is what usually is referred to as voltage drop.
extractor fans 2012/06/05 at 9:09 am #12971Spir Georges GHALIParticipant@sturnbull720 said:
Just wondering if anyone can help me, I have recently come across a large voltage drop issue when a client is using a mig welder. When the welder is in operation at full load current it draws 26 amps (single phase load). The voltage at the outlet at no load is 242V and under full load current is 208V. The voltage drop is due to a preexisting large run (over 50 metres) of 2.5mm2 twin and earth cable which is grossly under rated for its application. I am looking at upgrading the circuit cabling to larger CSA conductors which will overcome the voltage drop issue. My question is however, if the welder is drawing 26 amps at 208V, what will the current draw be at nominal voltage of approx 240V? I understand with a purely resistive load that, impedence remaining constant, as voltage decreases, current will do the same. I’m trying to avoid installing circuit wiring suitable for circuit length/voltage drop/load current, only to find out at nominal voltage the load will be considerably higher. Any information will help. ThanksDear ;
Normally, the voltage drop is from the source till the load, and the percentage at your case is around ” 14% ” ( too much ), so, is the measured value at the input points of welder or at the beginning of the 2.5mm² cable ? if at the input points of the welder, it’s better to measure it at the beginning of cable to decide if the changing this section will be enough or not.
2012/06/05 at 11:14 am #12973sturnbull720ParticipantThe no load voltage at the origin of the circuit is 242V and the no load voltage at the outlet where the welder is connected is 241V. With a total circuit load of 32 amps (26A for the welder plus another couple of small appliances at 6A) the voltage under load at the origin of the circuit is 237V and at the outlet where the welder is connected the under load voltage is 208V.
2012/06/05 at 4:27 pm #12975Spir Georges GHALIParticipant@sturnbull720 said:
The no load voltage at the origin of the circuit is 242V and the no load voltage at the outlet where the welder is connected is 241V. With a total circuit load of 32 amps (26A for the welder plus another couple of small appliances at 6A) the voltage under load at the origin of the circuit is 237V and at the outlet where the welder is connected the under load voltage is 208V.Dear ;
As you said that the voltage at the circuit’s origin with ” 32A ” is ” 237V ” ( the circuit origin is the place where the cable 2.5mm² is connected ), and at the connection points of the welder is ” 208V “, that means the voltage drop is caused by this cable, so :
– The distance is ” 50m “.
– The total load current is ” 32A ” where the welder current is ” 26A “.
As the welder current contains a lot of Harmonics that can’t be measured by the standard equipment, the real welder’s current is bigger than ” 26A “. For this case it’s better to use a cable’s section around ” 6mm² ” where the method of installation of this cable should be taken into consideration.

AuthorPosts
 You must be logged in to reply to this topic.