Author Topic: How is shorting supply +, - terminals different from joining them through load?  (Read 2628 times)

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Offline vineel567

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Hi All,

How is shorting a voltage supply +ve , -ve terminals difference from joining them through a load.
Why doesn't in the second case we don't blow the voltage supply(or spoil a battery).
Isn't joining the terminals through a load same as joining them through some load(apart from have load resistance)?

Thanks in advance.
 

Online BravoV

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Isn't joining the terminals through a load same as joining them through some load(apart from have load resistance)?

You've answered your own question at the bold section above without realizing your self.  ^-^

Joining .. shorting directly through PS terminals has really-really low load resistance because that cable is basically the load it self, while normal load or circuit, it has much higher resistance than just a piece of wire.

Use V=I R formula to rethink you question.

Offline AlfBaz

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How is shorting supply +, - terminals different from joining them through load?
One is a waste of electricity, the other is put to work... I'll let you figure out which is which
 

Offline vineel567

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Thanks alot BravoV, AlfBaz

So what actually matters in either case is connecting the supply with the appropriate resistance, This resistance can come from some load(bulb) or can also be through some thick wire(assume with very high resistance)  shorted.
In the first part of my question(shorting supply +, - terminals) because we are using very less resistance we are allowing huge currents to flow and chances of blowing the supply. Right?
So I can short a supply terminals provided the wire used has sufficient resistance(ideally). Is it correct?

Please correct me if my understanding is wrong.
 

Offline moepower

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Your supply will have a maximum current rating, or better yet, it may have a current limiting protection.   Let's assume that the supply is rated for 10 amps maximum.   You can use resistances across the terminals as long as the current does not exceed 10 amps.  You can work this out by V=IR.   R, resistance, represents your load.    The bigger the load, the smaller the resistance, the more current it draws.  An extreme case is the short-circuit, which is like an infinite load, where R approaches zero, and the current goes very high.

Does that make sense?  (This is for a dc supply I'm assuming)





« Last Edit: November 08, 2013, 02:37:20 pm by moepower »
 

Online BravoV

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.....  can also be through some thick wire(assume with very high resistance)  shorted.

You made a mistake there, the thicker the wire, the "lower" the resistance. Also shorter the wire has lower resistance than longer one.


So I can short a supply terminals provided the wire used has sufficient resistance(ideally). Is it correct?

If the real load for example your bulb has 10 Ohm resistance, and if you can find a piece of wire that also has the same 10 Ohm resistance, then both will have identical results.

A 10 Ohm wire is either very-very-very thin  :P , or either very-very-very long.  >:D


Note : Actually bulb resistance is not fixed, when heated the resistance will get higher, but for the simplicity sake, just assume it has a fixed resistance whether its on or off, otherwise this will confuse you even further.

Offline vineel567

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Thanks alot BravoV,

Your reply answered my questions. Thanks alot. EEVBlog is awesome for a noob like me :)
 

Offline Kremmen

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One more nit-picking comment: "shorting" and "connecting" should really be used as distinctly separate concepts. A "short-circuit" means that a path in a circuit is closed such that the resulting current (far) exceeds that which is designed to be the maximum allowable current of the particular circuit. In other words, you connect a load but you (accidentally) cause a short-circuit. There are exceptions like the crowbar circuit but you get my meaning.

My recommendation for the OP (vineel567) is to familiarize with the concept of current and voltage sources in basic circuit analysis. I mean the ideal voltage source and the ideal current source combined with their source resistances. Every practical source, such as the lab source in you question, can be seen as a combination of an ideal source together with its source resistance. Understanding the properties of ideal sources takes you a long way in understanding the answers to your question. My suggestion for this topic is to master the concepts outlined in the following Wikipedia link, and you will understand: http://en.wikipedia.org/wiki/Voltage_source

So, what exactly counts as a short-circuit? If you connect say a 0.05 ohm resistor as a load to your lab power supply (say a typical 30V, 3A supply like the one sitting on my workbench), the supply will reach its 3A current limit already at 150 mV of output voltage. For all practical purposes it can be considered a short circuit.
On the other hand, if that resistance happens to be the armature resistance of the starter motor in your car, the battery will happily supply ~200-250 amps to the load and crank up your engine. So in the latter case 0.05 ohms is just a load to the battery, not a short-circuit. A large lead-acid battery can have an internal resistance as low as ~3 milliohms, resulting in its ability to source huge currents for things like turning a starter motor. On the other hand, your lab supply can have an internal resistance of several hundred milliohms or even ohms. You won't notice that much because the lab supply is voltage regulated but still that is the way it is.
Nothing sings like a kilovolt.
Dr W. Bishop
 

Offline vineel567

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Thanks Kremman, I will follow the concepts in the Wiki.
 


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