What is more effective -- Current or Voltage?

[NARUTOSARTHAK]

I have fan that's 20 V and 14 A. I give it an input of 5V 1A. So my question is... increasing what will increase the performance of my fan more... Current or Voltage??

 

[atich]

That's a very broad question. It depends what you mean by performance. Speed, torque or efficiency? The speed of a motor is proportional to voltage and the torque is proportional to current. For maximum efficiency you need to operate at the maximum power point because efficiency = power out / power in So if you really want to find out the best place to operate you might have to plot a speed vs. torque curve. Needless to say you shouldn't exceed the voltage and/or current rating of the DC motor.

 

[sophiecentaur]

I have fan that's 20 V and 14 A. I give it an input of 5V 1A. So my question is... increasing what will increase the performance of my fan more... Current or Voltage??

You cannot dictate both values. The resistance of the load tells you the current it passes, with a given supply voltage. Because you are dealing, here, with a motor, it's not possible to predict how a fan that's rated at 20V (when it passes 14A), will behave with 5V across it (it's not as easy as a heater that will follow Ohm's Law). So your question is difficult to answer, as it stands. See aitch's post, above.
If you supply much more than 20V, the fan will probably go faster but could burn out.

 

[Dheeraj12]

Ohms law says v is prop to i , now if u v then how can u expect I to remain same.. So your question seems irrelevant to me

 

[Dheeraj12]

*increase .. Correction

 

[sophiecentaur]

now if u v then how can u

If you mean "you" then why not write "you"?
"u" could mean anything, in the context of Physics - possibly 'initial velocity' ?
And "Prop" is a position on the rugby field.
That's why we don't use txt speak on PF.

 

[Dheeraj12]

Sorry i meant if u (generally speaking) increase v then i will increase by ohms law

 

[sophiecentaur]

Not every device follows Ohm's Law.
It might help you if you read around this topic a bit. Usually, there is a positive proportionality between V and I but it is not always so. There are devices with Negative Resistance slopes in their characteristic. Start with plain resistance and move on when you have that sussed.

 

[meBigGuy]

In most situation the Voltage controls performance. Current is determined by what you connect the voltage to (assuming the supply is not a limitation).

Let's say we had 2 battery chargers (wall warts). One is 5V at 1A and the other is 5V at 2.1A. If we connect to a device that requires 5V at 1A, then all that will be drawn is the 1A required by the device. 2.1A is the maximum capacity of the second charger, that is, it can supply up to 2.1A and still output 5V. IF the load tries to draw 3A, then the output of the charger will likely drop out of specification, possibly to zero.

If a fan were strictly a linear device (I'm not saying it is), then cutting the voltage in half would cut the performance (power consumption) by a factor of 4, half due to the voltage decrease and half due to the resulting current decrease. But, be careful about that because fans are not linear.

Chances are when you connected your 5V 1A supply to a 20V 14A fan it essentially shorted the supply and the voltage went to zero or near it. The supply would probably burn out if you left it connected (if it didn't already).

I hpe u dcd 2 use real words. Rdg that bs is BS. dnt b so lzy if u wnt intelligent ans.

 

[sophiecentaur]

I hpe u dcd 2 use real words. Rdg that bs is BS. dnt b so lzy if u wnt intelligent ans.

Hr hr!