Electronics > Beginners
Some noob questions
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Mr D:
Thanks, it's getting ever so slightly clearer.

But now i've come up against the following stumbling block:

Say we have a simple circuit which is a 10v battery and a resistor.

If the resistor is 10 ohms, and we place our MM either side of the resistor, we measure 10v.

If we change the resistor to 5 ohms, and measure again, we still see 10v.

Ok, so 5 ohms is a worse resistor then 10 ohms, but we still measured 10v.

Let's make the resistor even worse. What's the worst resistor imaginable? No resistor!!!

Now if we measure either "side" of this non-resistor, we certainly won't see 10v on our MM.

So where does this relationship break down? I mean the relationship that says that voltages stay the same measured across a resistor, regardless of the resistor value?

james_s:
What do you mean "worse" resistor?

As you decrease the value of the resistor, it will draw more current from the battery which means it will dissipate more power. This breaks down when the resistance becomes so low that the source can't supply as much current as it is trying to draw and the voltage will sag below 10V. If you could have a 0.0000 Ohm resistor across it the voltage would sag down to 0 Volts but in reality you can never have quite zero.
rstofer:

--- Quote from: Mr D on July 19, 2018, 09:52:43 pm ---Thanks, it's getting ever so slightly clearer.

But now i've come up against the following stumbling block:

Say we have a simple circuit which is a 10v battery and a resistor.

If the resistor is 10 ohms, and we place our MM either side of the resistor, we measure 10v.

If we change the resistor to 5 ohms, and measure again, we still see 10v.

Ok, so 5 ohms is a worse resistor then 10 ohms, but we still measured 10v.

Let's make the resistor even worse. What's the worst resistor imaginable? No resistor!!!

Now if we measure either "side" of this non-resistor, we certainly won't see 10v on our MM.

So where does this relationship break down? I mean the relationship that says that voltages stay the same measured across a resistor, regardless of the resistor value?

--- End quote ---

As has been mentioned above in several posts, every voltage source will have an internal resistance.  For example, the internal resistance of a AAA battery is quite high when compared to a car battery.  You might survive putting 0 Ohms across a AAA but I would NOT suggest doing it to the car battery.

We know from Ohm's Law that the internal resistance will drop a voltage E = I * R.  It really is that simple!  The thing is, R is quite low so for a given E, I can be quite high.  Like the car battery - it has a very low internal resistance.

This is the Thevenin Equivalent Resistance I mentioned way back near the top and several others have mentioned.

When you put a light load (1A (10 Ohm 10V) or 2A (5 Ohm 10V) on a source with sufficiently low internal resistance, you will measure essentially the full source voltage.  When you put a very low resistance across the source, the internal resistance dominates and the majority of the voltage is dropped inside the device.  Take that internal voltage drop V and square it.  Then divide by the size of the resistor to get the power dissipated in Watts (E squared over R).  Notice how a resistor less than 1 Ohm drives the power up pretty quick.

That's also why I brought up the voltage divider experiment above.  I was using reasonable resistors such that internal heating of the source wasn't likely to occur.  But the top resistor does exactly the same thing when you slide it inside the battery and call it the Thevenin Equivalent Resistance.  The internal resistance forms a voltage divider with a very low value for the top resistor.

Later on you will find that maximum power transfer to the load occurs when the load resistance is equal to the internal resistance.  This is a consequence of the voltage divider equation given above.

If the external load resistance is higher than the internal resistance, the majority of the voltage drop is outside the battery.  Think 1 Ohm internal resistance and 10k load resistance.  Essentially all of the voltage appears across the load resistance.  Now turn it around:  Think of the 1 Ohm internal resistance and a 0.1 Ohm load resistance.  In this case only 10% (approx) of the voltage drop is at the load and the internal resistance is dropping 90% (approx).  And you can bet the battery is heating up.

You can't get around the necessity of wrapping your head around Ohm's Law.  It explains everything but you need to play with it to get an intuitive feel.  What happens to I when you vary E for constant R.  What happens to I when you vary R with constant E?  What happens to E when you vary R with constant I?  Sit down with a scratch pad and lay down that voltage divider and see how the number actually work.  Same with the voltage divider using the battery internal resistance (assume 1 Ohm just for a number) and various external resistors or even an external 2 resistor divider.

james_s:

--- Quote from: rstofer on July 20, 2018, 12:07:40 am ---As has been mentioned above in several posts, every voltage source will have an internal resistance.  For example, the internal resistance of a AAA battery is quite high when compared to a car battery.  You might survive putting 0 Ohms across a AAA but I would suggest doing it to the car battery.

--- End quote ---


I'm gonna guess you meant to say you would *not* suggest doing that to a car battery. The results are likely to be fairly spectacular but certainly not something I'd recommend trying.
rstofer:

--- Quote from: james_s on July 20, 2018, 12:17:39 am ---
--- Quote from: rstofer on July 20, 2018, 12:07:40 am ---As has been mentioned above in several posts, every voltage source will have an internal resistance.  For example, the internal resistance of a AAA battery is quite high when compared to a car battery.  You might survive putting 0 Ohms across a AAA but I would suggest doing it to the car battery.

--- End quote ---


I'm gonna guess you meant to say you would *not* suggest doing that to a car battery. The results are likely to be fairly spectacular but certainly not something I'd recommend trying.

--- End quote ---

Fixed!  Thanks...
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