# B Electricity, voltage, current, resistance relationship

1. Apr 3, 2017

### Kajan thana

Hi,
The Circuit I am referring to is a simple series circuit with the resistance of the wire is negligible.

I am confused about the fact that if we increase the resistance of a resistor in a circuit, the PD across the component will also increase. From my understanding I thought increasing the resistance of the component will increase the overall resistance so the overall current will decrease. If the voltage is proportional to current, the the voltage should increase; this explanation seem to be wrong.

I tried to search this up on google, didn't get a proper answer, can someone explain this to properly please ( without the water analogy) ?

2. Apr 3, 2017

### Staff: Mentor

It just depends on what kind of a source you are using to drive the resistor string. If you are using a voltage source, the total PD stays the same as you change the load resistors, and the current changes via I=V/R.

If you are using a current source instead, the overall PD changes via V=IR. Does that help?

3. Apr 3, 2017

### Kajan thana

Slightly, I came across a question where the source is voltage but as the resistance increase, the voltage across the component also increases, I used some sort of animation ( virtual circuit) , it proves that the voltage is proportional to resistance but the current decreases. I have attached to two screenshot for a better understanding of my question. I don't understand the proportionality. I thought one element need to remain constant for proportionality to work, but in this case all the three elements varies.

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4. Apr 3, 2017

### Staff: Mentor

I'm not understanding the figures. Where is the voltage source? Is the current measured at the bottom done with an ammeter? Or is that a current source of some kind at the bottom? Could you post a more traditional schematic of what you are simulating?

5. Apr 3, 2017

### Kajan thana

I have attached the screenshots again with some annotation. Thanks.

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6. Apr 3, 2017

### Staff: Mentor

The pictures that you posted show 9 V across the component in both cases. I am not sure what you mean by "the voltage across the component also increases". The variation in the least significant digit appears to be just rounding.

7. Apr 3, 2017

### Kajan thana

In the screenshot, it shows two different voltages ( 4 S.F) before increasing- 8.997 After increasing- 8.999

And if I further decrease the resistance, it will show 8.991

8. Apr 3, 2017

### Khashishi

In general, there is some internal resistance inside a realistic voltage source. You can model this as a ideal voltage source in series with a small resistor. So, you have a voltage divider between your external resistor and a small resistor in the battery.

Also, you have a large but finite resistance in a real voltmeter, and a small resistance in a real ammeter. I don't know if any of these effects are modeled in your simulation, but it can explain your results.

9. Apr 3, 2017

### Staff: Mentor

That appears to be just rounding error in the least significant digit for whatever simulator you are using. Just ignore it.

To put this in perspective, you have changed the current and resistance by a factor of more than 2 and the voltage has changed by a couple parts in 10000. That is essentially constant.

10. Apr 3, 2017

### Kajan thana

I have attached a ideal circuit where the internal resistance and resistance of the wire is negligible .
If I say the temperature of of the thermistor is constant, then why does the voltage across R increase when we increase the Resistance of R. I am still misunderstanding the concept.

(NOT A HOMEWORK QUESTION)

Thanks

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• ###### Screen Shot 2017-04-03 at 17.39.41.png
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11. Apr 3, 2017

### cnh1995

What happens to the "total" resistance if you increase the value of R?

12. Apr 3, 2017

### Kajan thana

The total resistance will also increase as they are in series. so the current will decrease.

13. Apr 3, 2017

### Khashishi

14. Apr 3, 2017

### cnh1995

Yes.
But since the thermistor resistance is constant, what can you say about the voltage across it?

15. Apr 3, 2017

### Kajan thana

It will decrease

16. Apr 3, 2017

### cnh1995

Yes. So what can you say about the voltage across R?

17. Apr 3, 2017

### Kajan thana

So the other one have to increase. Thank you. Make sense now.

18. Apr 3, 2017

### Kajan thana

Thank you

19. Apr 3, 2017

### davenn

This circuit is totally different to your first one !
you should stick to one circuit at a time and learn to understand it, before making changes

you are measuring the voltage drop across that thermistor.
The voltage drop will vary as the thermistor changes temperature and therefore changes resistance
if kept constant temp, then the voltage across it wont change
If you vary the resistance of R, the voltage will change across the thermistor and R

how do you know it changes ( yes it does) but you are not measuring the voltage across R

IF you put your voltmeter across R and change the value of R, then yes the voltage will change according to Ohms Law
why, because the current through it will change

you could plot a graph of V for different values of R and I (current)

NOTE: If R was the only resistance in the circuit ( the thermistor or other resistance wasn't there)
Then the voltage drop across it wouldn't change !. Only the values of I and R would change as
the value of R was adjusted

Dave

Last edited: Apr 3, 2017
20. Apr 4, 2017

### PeterO

If you have the simple "two resistors in series" we have what is called a voltage divider circuit.
If the two resistors are equal, the provided Potential is divided equally. Perhaps a "6 Volt battery" resulting in a Potential Difference across each resistor of 3V.

If you were to increase one of the resistors (lets say double it) the size of the resistors would now be in the ratio 2:1, so the applied Potential is divided in the ratio 2:1. If that same 6 Volt battery is applied, that means the Potential differences will now be 4V and 2V.
So the increased resistor now has a larger Potential Difference across it (or perhaps you want to call it voltage drop - but most people unfortunately just call it the voltage).
Interestingly, in the second case the PD will be higher (4V vs 3V) but the current will actually be reduced. This may sound counter intuitive, but remember the Potential Difference is calculated using Ohms Law: V = I x R
If you Double R and halve I, V will be the same. But if you double R, but only reduce I to two thirds of the original, the value of V is larger.

Note: While it is common to refer to resistance, current and voltage, it is much better to refer to "resistance of", "current through" and "potential Difference across" any part of a particular circuit. "Voltage Drop across" is a poor, but acceptable, alternative to "Potential Difference across" but to simply say "voltage" is to be avoided.
Note: Potential Difference is measured in Volts, and everyone wants to call it voltage. A person's weight is measure in kilograms (or newtons if you insist) but no-one tries to call that kilogramage or newtonage. Why are they so "lazy" with Potential difference.

21. Apr 4, 2017

### sophiecentaur

From what you say, I get the impression that you may not have done much basic Electronic Theory and that you're wanting an answer without using it. Your question seems to refer to a very elementary circuit configuration, referred to as the Potential Divider. This is just the second or third step on the ladder of conventional learning of Electronics. Maybe you have not recognised the circuit for what it is. If you start from the very basics of resistive circuits and work upwards, you should find all this stuff pretty straightforward with no mystery about it at all.
This may have been pointed out earlier in the thread but I am just restating it in louder terms. Simulations (in general) are the cause of a lot of elementary problems for people who want to rely on them, rather than to get some basics under their belt first. Simulations are the product of Theory and not the other way round.

Last edited: Apr 4, 2017
22. Apr 5, 2017

### Kajan thana

Thank You Dave.

23. Apr 5, 2017

### Kajan thana

Thank You

24. Apr 5, 2017