Load Current

This looks like a Homework question, so I'll give you a Home work answer.

Look to Ohms Law for your answer. and convert your HP value to a kW value to make your life easier.

Regards,
Sapper

Hello Sapper,

1 HP = 745.69987158227022W.

Please pay attention to my question. The question was, in nut shell, the load remaining the same, whi did the current droped when the applied voltage increased?

Can you give me an inteligent answer? Otherwise please keep quiet.

Zodiac,

I did.

Power is a function of Voltage (V) and Current (I) in Watts (W).

If you did your homework yourself you would understand that by multiplying V*I for each set of figures the Wattage is almost the same.

Regards,
Sapper

Friends, friends, lets keep our cool, that's the only way to learn

Yes, V*I is required to be the same since it is a motor, and NOT an ohmic load only. So, when given a higher voltage of 223 V, rather than drawing a higher current, it only draws enough to drive its load mechanically, so W is constant.

Would you like to consider Power Factor also, given that this is AC?

It is a principle of electricity!! Just like transformer law- when voltage is low current is high and when voltage is high current is low

When there is a short in the system circuit that is when load draws high current from supply irrigardless of voltage level.

Thanks,

Joseph

Just think how low your current would be if you put 50,000 Volts through the motor!!!

By the way, I know a good transformer lawyer if you need one.

By basic electrical engineering calculation, power is alway constant for any electrical equipment while the voltage and current may vary.

if the voltage vary at constant power, then the current will also vary.

i.e P =IV

I = P/V

The larger the voltage the smaller the current, all in a bid to equalize the Power.

The only problem is when a voltage smaller than what is equires to power such equipment, it will take extra time tp produce the needed current and there causing a heating effect on the circuit.

I am a little confused here. Suppose you take the case of a resistance furnace or an incandescent bulb, will the current not increase with an increase in voltage ? power will actually increase in these cases, since these will purely follow Ohm's law.

It is only in the case of motors that if the voltage drops, the current will increase to compensate for it so that the mechanical output is maintained (within limits of course). The slip will increase reducing the back EMF. If the voltage drops too much, the motor will stall.

Dear kvsridhar,

It makes sense. I believe that the 'slip' causes to slow down the motor and there by reduce the back EMF which oposes the supply current. Thanks for the refreshment. I had degree in mechanical engineerig with some expossure to electrical engineering, 39 years ago and naturally did not remember these and lots of other stuffs. I believe that the forum is to help one another and you helped me remember the forgotten.

Just one more question to the forum; Suppose you increased the voltage gradually so that the motor reached its synchronous speed. At this speed the current must be the minimum. What happens if you increase the voltage further? Will it not increase the current as you increase the current?

Hello Sapper,

People those ask questions may not necessarily be students asking help with home works. I have done my home works 39 years ago. Intelligent answers may benifit others, possibly more than me, in the forum. So please be helpfull being silent if you are unable to give correct answer.

Thank you Shridhar.

Thanks for your nice sentiments. At 65 years of age, my electrical fundamentals are barely visible in the mists of time. However, as I am still active as a consultant, some of it remains.

Firstly, an induction motor can NEVER reach synchronous speed. Why? Because at synchronous speed, the torque developed is zero, the roror having no relative movement with respect to the stator's rotating filed. So, induction motors HAVE to have a slip. The more the torque demanded from the motor, the more the slip.

Secondly, yes, you are right in that the current will increase if you increase the voltage beyond the rated voltage.It will work somewhat similar to an ohmic load, and it will overheat. If there is no thermal protection, it may well burn out.

Hope this helps.

may I create a little bit more confusion? .

Motor (Induction or otherwise) is one which will convert electrical energy into mech energy.

hence the power requirement by the motor is the one demanded by the load.

In a certain type of load - as the voltage increases the current drops and vice versa- where the demand is almost a constant mechanical power. But not always

There are motors that will not demand more current as voltage drops, it may be happy (almost) with what it gets.

The Idea is the motor characteristics intersection with the load characteristics define the different power demands and that defines the operating point.

Voltage changes → motor speed = f(v) changes → load KW = f(speed) changes→ Motor Input KW = F(load KW, efficiency at the speed) changes → current =f(V,KW) changes.

This is due to the non-linear characteristics of the load.

But if you have the ideal linear load - pure ohmic resistance - almost say your incandescent bulbs, the power is f(V²) = V²/R so the power changes in that direction as voltage changes with linearity in I too Since I = V/R so Voltage change changes current linearly and power accordingly.

Note again (stress ) the motor may be like that but not the load on the motor.

And motor will not work like ohmic- Go above rated voltage, the current will not increase, as it reaches the near synch, it can not go further, the power (asssuming no load say) will remain same so current will keep on reducing - Even if there is a load, as it reaches near s=0, (of course it can not reach the asymptote, just be nearer and nearer to it) the speed stabilises, so does the load, so current will keep on dropping- and it will never behave to increase the current.

I must beg to differ regarding the current going down as voltage goes above rated. Here is a page explaining what happens...saturation is one of the differentiators

Yes you are right, since the motor might reach the saturation portion of the Hyst curve and draw the extra reactive current. The above excerpt provided was from cowern papers ?

http://www.motorsanddrives.com/cowern/motorterms12.html

In fact if one goes to the home-page, there are many simply explained info on motors.

http://www.motorsanddrives.com/

Absolutely right. Yes, the home page has lots of good stuff. I am no expert of any kind on motors, but being a switchgear and controlgear designer, had to know some minimum stuff about the loads which my products had to operate, control and protect. Thank you.

I'm getting confused by names here! Are kvsridhar and Shridhar the same person?

If anybody's interested, I came with a formula relating motor speed to voltage, assuming, among other things, that back-EMF = applied volts at synchronous speed, and varies directly with speed.

ω_V2 = (V₂² _* ω_s – V₁² _* (ω_s - ω_V1)) / V₂²

^{where V}1 ^{= voltage 1, V}2^{= voltage 2, ω_V1 = speed at voltage 1, ω}V2^{= speed at voltage 2, ω}s ^{= synchronous speed.}

Agrees pretty well with published motor data.

As the speed increases slightly with increasing voltage, load power also increases, so the current falls a bit less than calculated from voltage ratios. In principle, the load could be modified to maintain constant power, but in practice it varies as speed³ for a squared-torque load, directly with speed for constant torque load.

Cheers........Codey

Hi Codemaster,

Yes, this being India, we value people more than names and Sridhar, Shridhar can be used interchangeably Pardon me, no affront is meant, it is just a tongue-in-cheek about our cavalier-like attitude to spellings !

Whats in a name after all as Shakespeare said ...

Your search result is fabulous. It is right on the button.

However, in my realm as a designer of motor starters, i have dealt predominantly with 3 phase induction motors, and they behave as per the speed-torque characteritics as published by the manufacturer. I can publish a typical characteritic if required, but it is premature for this thread i think.

When the voltage is low, the motor will draw a higher current. It will need to do so, since it has to deliver the power required by the mechanical load. Its speed will drop since the slip has to increase to deliver the torque required.

When the voltage is equal to rated, it will draw the rated current, and everything is hunky dory.

When the voltage exceeds the rated voltage, the current drawn goes up, and the winidngs will overheat, If the voltage is so high that the current drq\awn is higher than the overload setting of the protectove relay, it wil trip and protect the motor. If there is no relay ...? Motor will burn of course.

With me so far?

Hello kvsridhar

Thanks for the clarification!

It was the speed-torque (hence power) characteritic of the load I was talking about, not the motor. You're right, these relationships only work within a reasonable range of voltage, I believe manufacturers say +/- 5% of nominal. Probably OK in practice a bit outside that, but if you go too far things will go awry.

Cheers.......Codey

Fantastic!!

And thank you for the input Codemaster.

As for "Shridher", it is a very popular name in India. I assumed that "KV" must be initials. Usually the first letter stands for the initial of the family name (Last name) and the second letter satands for the initial of father's first name (Middle initial of the person). In Kerala, (A suthern state of India) the name will be written as "K. V. Shridher". Kvshridher, correct me if I am wrong.

Best regards.

Hi Codemaster,

Sorry i did not realise you were referring to the load characteristic. You are quite right of course.

And Zodiac.

Yes, my name is indeed K V Sridhar, with the initials as you describe. The given name is as spelled here.

Regards.

Here the percentage impedance of the transformer plays very important role.Suppose your 200VA transformer is having an impedance of 5% ie, if you apply the full load the secondary voltage across the terminals will be 236*5/100 ie, 11.8V less from 236V that means the voltage will be 236-11.8=24.2V. It does'nt matter whether the load is inductive,capacitive,resistive and the combination of all.Your second source probably have less impedance than the first source.No information about the rated voltage,rated current,frequency of the fan which can be collected from fan data plate or from manufacturer's specification .Apply the very same voltage to fan terminals to get the maximum efficiency @ rated torque.check for rated current and analyse the losses according to variations in measured value.

Moter is 1/4 HP means 186 watts.

What is the power factor of transformer? If the power factor is 0.8, then your transformer rating

200 VA is 160 watts

250 VA is 200 watts

When you run motor on 200VA, its voltage drop due to overlading of transformer. When a transformer is overloaded, its copper losses inreases. Current on motor is higher due to drop of votage.

ON 250 VA transformer which is 200 watt less voltage drop resulting less ampere.

As we all know that

P=VI Cos Ø for singel phase

P=√3 VI Cos Ø for 3 phase

where

P= Power in watts

V= Volt

I= Ampere

Cos Ø = Power Factor