Voltage and Current

Q1) Is it possible that a machine burns out due to voltage drop at its terminal? I mean that the supply voltage is not equal to the rated voltage=85 Q2) Is it true that for a constant KVA load current increases as the voltage decreases? What is a constant KVA load? (Example) We know that current and voltage are directly related. As voltage increases current increases as well. Can any one explain?

Reply to
Ali
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Ali schrieb:

Yes, a machine can burn out due to temperature excess at its terminal caused by bad conduction. The voltage drop in conjunction with the flowing current dissipates power at the terminal and rises the temperature.

A load driven by a SMPS. The output power will be constant while the product of input voltage and input current will be nearly constant.

- Udo

Reply to
Udo Piechottka

Ali wrote in news:80e4bfc0-999d-487d-b028- snipped-for-privacy@y21g2000hsf.googlegroups.com:

Voltage and current are inversely related. As voltage increases, current decreases, vice versa.

Reply to
HaLiGoOn

#1 Yes it's possible to burn out certain devices with low line voltage. Most however would NOT burn out, may trip a breaker or just run poorly. It all depends on what the device is and its particular characteristics.

#2 If it is truly constant KVA, and not constant current or constant voltage or some other regulation, the load current and voltage would be inversely proportional, one goes up, the other goes down.

If the load is resistive or simple reactive, not constant anything; e.g., a light bulb, the voltage and current are directly proportional, one goes up, the other goes up as well.

In otherwords, the voltage-current relationship depends on the device.

Reply to
Bob Eld
  1. Yes. Motor.

  1. Yes. Motor.

Reply to
operator jay

i did an experiment on single phase fan. using auto tarnsformer i was controlling the voltage. the power output decreases as well as the current. i want to know, that which devices exactly follow the inverse principle. i.e. when voltage decreases current increases

Reply to
Ali

You have been given one example - switch mode power supply units.

Here are some more: * plasma (plasma channel) * electric arc* * fluorescent lamp * tunnel diode * IMPATT diode * unijunction transistor * resonant tunneling diode * resonant tunneling transistor

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If you explain why you want to know, it might help. No one is going to list every single device that works in this way.

-- Sue

Reply to
Palindrome

i did an experiment on single phase fan. using auto tarnsformer i was controlling the voltage. the power output decreases as well as the current. i want to know, that which devices exactly follow the inverse principle. i.e. when voltage decreases current increases

Some motors with a friction load will begin to lock up and draw more current when the input voltage is insufficient to keep the rotor running.

peace dawg

Reply to
We Can Do It

i did an experiment on single phase fan. using auto tarnsformer i was controlling the voltage. the power output decreases as well as the current. i want to know, that which devices exactly follow the inverse principle. i.e. when voltage decreases current increases

----------------------- The fan load torque will decrease with speed causing the overall current and power to drop. A 5HP motor for example will draw more current at a lower voltage when operating near full load but will draw less current at start. The behaviour is not only dependent on the voltage but on the behaviour of the load being driven. In your Q2 the KVA is constant and so I must decrease when voltage increases. (VA=Volts*amps).

Ohms law doesn't apply to motors or to constant KVA loads.

Reply to
Don Kelly

A. Yes.

It would be unusual in the real world for the load to actually remain constant at the reduced speed, but in the hypothetical world of this homework problem, it can be assumed. The approximation is conservative, and usually close enough.

At a constant load, a motor will turn slower at a reduced voltage, and the current drawn will increase. This is usually explained in terms of a quantity called "back EMF", which is an assumed voltage generated by the rotation of the motor, and which opposes the applied voltage. The faster the motor turns, the greater the "back EMf".

Because power dissipated is proportional to the square of the current, the motor will become hotter at the reduced voltage. The temperature may or may not reach a level high enough to actually melt (i.e. burn out) the motor windings.

Anthony Straight tonyelectric.com

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Reply to
Tony

A good example is burning out an electric chain saw motor when used with a long extension cord.

They all have warnings not to use with cords below a given gauge and length.

Reply to
Rich256

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