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Potentially naive question electric circuit alert
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Minky
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Post Potentially naive question electric circuit alert  Reply with quote
Okay, I'll dive in..  Why, when the N-channel Mosfet in the bottom left of the below diagram activates, doesn't the pack short?



Do resistors cause work for the current? And therefore the wire from the resistor is effectively a negative?

I don't know why I started thinking about it but now it is really bugging me.

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Mon Feb 13, 2017 10:43 pm View user's profile Send private message
SSGT
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Resistors resist current flow according to Ohm's Law i.e. V=I*R (or, in this case, I=V/R).

The current flowing through a circuit is a function of the voltage and the resistance. If you have a voltage of 12V across the resistor then only 0.0012A (1.2mA) of current will flow through it (I=12V/10,000Ω). A short circuit is a low resistance connection between the battery terminals. Lower resistance = higher current for a given voltage.

The FET will also introduce some resistance (normally a few milliohms "on" resistance referred to as RDS(ON) i.e. resistance between drain and source when on) although we say it's effectively zero because an additional few milliohms is negligible when compared to the 10,000 Ohms of the resistor (the difference between 10,000,000mΩ and 10,000,015mΩ just doesn't matter especially when that "10kΩ" resistor could actually be anywhere from 9-11kΩ or 9.5-10.5kΩ depending on the tolerance bands). If we follow this assumption that means the voltage at the bottom of the resistor will effectively be the same as at the pack -ve terminal and the voltage at the top of the resistor will be the same as the pack +ve terminal. It's not good to think of it becoming a negative terminal but it will be at the same voltage as that terminal.

The key to remember is that absolute voltage rarely, if ever, matters - we're almost always interested in the voltage between two points (i.e. the "potential difference"). If, for instance, you separate that resistor into two 5kΩ resistors wired in series the current flowing through them would be the same but the voltage in between the two resistors would be half the pack voltage (i.e. each resistor will have 6V across it). This makes sense because 6V/5,000Ω (the current through each individual resistor) is the same as 12V/10,000 (the current through both) as they are in series. You could break that 10kΩ resistor into 10 1kΩ resistors and the effect would be the same (each resistor would only have 1.2V across it and 1.2V/1000Ω=1.2mA of current) or even 10,000 1Ω resistors (each resistor would only have 1.2mV across it and, again, 1.2mV/1Ω=1.2mA of current). It gets more complicated when you have different resistances in series but Ohm's Law always holds.

It's also worth noting that you could, if you wished, add a resistor in the diagram across the motor terminals to represent "air" which has a measurable resistivity and therefore, for a given terminal spacing, resistance. Electrical current can pass through air but air has such a high resistivity that you also need a massive voltage (and/or a very small gap) between the two points to do so meaningfully. Obviously air doesn't short a battery because the resistance of the air between the two terminals is ridiculously high (for most intents and purposes you can assume infinite) but it is there and is technically a potential path for current to flow.

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Last edited by SSGT on Mon May 08, 2017 4:33 pm; edited 1 time in total
Mon Feb 13, 2017 11:26 pm View user's profile Send private message
Minky
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Post Reply with quote
Thanks SSGT. I'm embarrassed to say that I thought that there was some fundamental quality change in the Electrons that had been 'used' to work.  I feel as daft as the guy MTB mentioned who thought there was an actual quality difference between red and black wire.

So a short is when you allow the full pack voltage easy access back to the pack and this is prevented by either constricting or stopping it's current/voltage through resistance or dissipating it's voltage through 'work' (converting it's energy into light, heat, volume, etc.), does that sound right?
If the above is true then why do we need the same gauge wire on the opposite side of a motor for instance?

When a battery shorts does it draw upon itself causing super high current through the intervening parts?

I'm really sorry, I know enough to know this is really, really, really basic stuff that I had thought I'd covered off in some background reading back when I started this. If anyone can point me in the direction of a electricity for idiots to save me wasting more of SSGTs time I'd appreciate it. Bearing in mind I'd read GCSE stuff and not got it.

Note: Even though I have to admit this is causing me great deal of frustration at myself I think it's important for anyone on here to keep asking the stupid questions. There will be someone else in your boots and although I've learnt a lot by asking my own I've learnt more by reading others and their answers. It's not like I'm a Nerf world council elder or something, I just think it's important because half the time (like me above) you don't even know you have a question to ask.

EDIT: I'm sorry I got my martyr on this morning with my whole 'I'm stupid for your sins' bit.  Nope. Just ignorant.
Tue Feb 14, 2017 8:44 am View user's profile Send private message
SSGT
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Post Reply with quote
Minky wrote:
So a short is when you allow the full pack voltage easy access back to the pack and this is prevented by either constricting or stopping it's current/voltage through resistance or dissipating it's voltage through 'work' (converting it's energy into light, heat, volume, etc.), does that sound right?


Effectively, yes. Resisting current flow through resistance and dissipating electrical energy to heat is effectively the same thing - a resistor resists the flow of current and the energy required to overcome the resistance is given off as heat. It's not technically "work" though unless that energy is used to physically move something.

Minky wrote:
If the above is true then why do we need the same gauge wire on the opposite side of a motor for instance?


Current is simply the rate of flow of electric charge (usually, but not exclusively, electrons). You can never have more electrons entering a point than you can leaving - the "flow" has to be continuous. You don't actually "lose" current in a resistor or motor but you do "lose" voltage through it.

A good analogy is a closed water pump circuit (things that you can visualise tend to be easier to understand). The water represents the electrons and is always "there" it just isn't always moving. The pump is effectively the battery and the pressure it generates between inlet and outlet is the voltage (or potential difference or electromotive force/EMF). Any restrictions to flow are effectively resistance and a turbine would be the equivalent of a motor. The pressure (voltage) generated by the pump (battery) forces the water (electrons) to flow (current). If you have no restrictions (resistance) then, assuming the pumps max flow rate (battery's max current) isn't a limiting factor, the flow of water (current) would be extremely high (assuming no losses it would be infinite as would current in a short scenario as 12V/0Ω=∞A - in reality, in both cases, the pipe/conductor will pose some resistance and the pump/battery will only be physically capable of a limited flow rate/current, at least before being damaged). If you add a flow restriction (resistance) the pump (battery) can't allow water (current) to flow as fast as it could unless you increase the pressure (voltage) across it - you can't physically have more water (current) flowing into a point than you have leaving it so the flow through the whole circuit is reduced. If you add a turbine then the flow (current) will cause it to spin and produce work - in both cases, restriction (resistor) and turbine (motor) the pressure (voltage) at exit will be less than at inlet and will be (assuming no pipe/conductor losses) the same as at the inlet to the pump (-ve battery terminal). It's not a perfect analogy but it's useful to illustrate the conservation of current aswell as a few other things.

Minky wrote:
When a battery shorts does it draw upon itself causing super high current through the intervening parts?


Again, it's better to think of the battery simply providing a set electromotive force that it applies to the electrons. If you have no resistance to oppose that force then the current will flow very rapidly. The current isn't created per-se, the electrons are already there and the battery simply forces them around the circuit from one terminal to the other.

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Last edited by SSGT on Wed Feb 15, 2017 1:58 am; edited 1 time in total
Tue Feb 14, 2017 9:53 pm View user's profile Send private message
Minky
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Post Reply with quote
Thanks SSGT. Apologies for wasting your time on very basics, your explanation is actually close to how I thought it worked.  I hadn't understood that the resistor was effectively dropping the voltage on the other side of it in exactly the same way as a motor or other device would do. This despite resistoring all those LED circuits down to 3v!  A moment of brain fart.

Thanks again.



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