How Does Adjusting a Rheostat Impact Voltage Distribution in a Circuit?

[kira506]

I'm talking about the primitive rheostat with coils and a slide and such , why when we inc. The resistance of rheostat does Vext inc and Vint dec. ?

 

[sophiecentaur]

If you draw the (complete) circuit, it is fairly easy to predict what will happen and give a reason. A verbal description may not be supplying enough information.

 

[kira506]

Yes,I've tried but I just can't seem to grasp a full idea of how V-internal decreases (which is equal to I r small) because current decreases when resistance of rheostat inc , and V-external increases whish is equal to IR , I mean the current is the same as i both ! Unless the current is different ?

 

[sophiecentaur]

This is a bit like trying 'grasp' how the supermarket bill works. It's down to the sums and the very basics of electricity more than anything. You will be able to feel that you have a grasp when you follow the sums through.
Have you drawn the circuit?
Mark the rheostat as R1 and the Load as R2. If Vin is the supply voltage then
The total current I through the two resistors will be Vin/ (R1+R2)
This current is, of course, dependent upon the two resistor values.
You say:

("Unless the current is different

Of course it's different! I = V/R always and it's that total series resistance that counts.
Clearly, I will get greater as R1 goes down (i.e. as the total resistance decreases).
So the current through the load will increase (Same current round the whole loop)
That means that the volts across the load Vout = IR2, will increase because I has increased.
Naturally, if the volts across the load have increased, the volts across the rheostat will have decreased because there is the same total supply voltage across the two resistors and it's the sum of the two volts.
The smaller resistor gets the smaller share of the total voltage.

 

[kira506]

But whenever I try it out , I find that by decreasing the total current ,the total voltage increases (decreasing total current by increaasing resistnance of rheostat) and there's this other resistor connected in series to the rheostat , the voltage in the second resistance (which is fixed) decreases by decreasig current , how can the total voltage increase while part of it decreases , please help me,the answer's like that in my textbook and I can't fully interpret it

 

[kira506]

This is a bit like trying 'grasp' how the supermarket bill works. It's down to the sums and the very basics of electricity more than anything. You will be able to feel that you have a grasp when you follow the sums through.
Have you drawn the circuit?
Mark the rheostat as R1 and the Load as R2. If Vin is the supply voltage then
The total current I through the two resistors will be Vin/ (R1+R2)
This current is, of course, dependent upon the two resistor values.
You say:

Of course it's different! I = V/R always and it's that total series resistance that counts.
Clearly, I will get greater as R1 goes down (i.e. as the total resistance decreases).
So the current through the load will increase (Same current round the whole loop)
That means that the volts across the load Vout = IR2, will increase because I has increased.
Naturally, if the volts across the load have increased, the volts across the rheostat will have decreased because there is the same total supply voltage across the two resistors and it's the sum of the two volts.
The smaller resistor gets the smaller share of the total voltage.

XD thanks a lot , you imagined the circuit with me and explained it quite smoothly , well I've got another question,the one that confused me the most , I'll base it on the circuit you imagined , the Q : what happens to Vload and Vout on increasing R1 ? And find a relation between V load an Vout where Vout = Vint (total of battery)- Ir where r small is the internal resistance