# Basic question on resistance - seeking a better understanding

1. Feb 20, 2013

### kjamha

Let's say a AA battery (1.5V) is hooked up to a simple circuit to power up a bulb (I want to assume the bulb lights up). If I switch out the AA battery for a D size battery (1.5 V), the light bulb will be brighter. Because the V is the same, I assume the current went up. Ohms law says V=IR therefore R=V/I so if I goes up, then the resistance goes down. I see resistance as the hindrance of the flow of charge. It is confusing to me that the increase in current will mean a decrease in the hindrance of the flow of charge. Can someone point out the where I am going wrong - or maybe provide me with a better definition of resistance? Thanks!

2. Feb 20, 2013

### Crazymechanic

Just because the current goes up or is increasing doesn't mean a part of a circuit or a resistor will have a lower resistance.
In fact in most cases when the current increases the heat dissipated by the resistor or lamp increases so the conductor or resistors temperature increases and the resistance rises.
So quite the opposite is happening.
Why do you think they use all that liquid nitrogen and other coolers to keep the superconducting magnets and wires cool? Because heat increases resistance rather than decreases if it would decrease the resistance by increasing current then ...victory :D:D Nop it's not that way.
There are some applications in which increase in current further decreases the resistance and further increases the current like in a gas discharge lamp.It has a positive feedback loop the more current the hotter the gas the more it conducts the more current is able to go through, so they use inductors as current limiters (they only work for AC) but that's not your case i said this for just the fact.

Last edited: Feb 20, 2013
3. Feb 20, 2013

### nasu

The D battery may have a lower value of the internal resistance.
In a circuit with low resistance, this factor may be important.

But is this really something you have observed?
Or you just assume that it will be like this? (the bulb being brighter).

4. Feb 20, 2013

### kjamha

Crazymechanic, you are confirming my confusion. I would expect R to increase. But V=IR, and if V remains constant it does not make sense that both I and R increase.
Also, thank you nasu. Two points: I did not consider the internal resistance of the batteries, but I would think the D battery has a higher internal resistance than the AA. Also, I did not try to hook up this circuit, but I thought the difference between a AA and a D is that the D has the ability to provide more current (at least that is what I have read). Because P=IV, I am assuming the I goes up and the Power is higher (brighter).

5. Feb 20, 2013

### Naty1

What do you say this? What leads you to this conclusion?

If you assume an ideal voltage source, 1.5 volts, the bulb will NOT be brighter with a 'D' battery.
But if you want to study such a scenario of different power sources in more detail, you have to consider this:

It IS important! [It may be higher internal resistance or lower, I do not know.]

If you want to properly model a voltage source, see here:
http://en.wikipedia.org/wiki/Thévenin's_theorem

With 'two resistances' now in your circuit, you can tell what will happen with different size internal battery resistances for a fixed resistance load.....

You will find the relationship may not be what you would think:
http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

Read carefully, this takes some getting used to....

6. Feb 20, 2013

### kjamha

I just built a simple circuit with a flash light bulb and a AA battery and there was very dim glow. I hooked up the D battery, and the glow was very bright.

So now I am back to my original confusion. The voltage was the same and I imagine the D battery provided more current, and therefore more power (P=IV). What is confusing me is ohms law. If V=IR and V is constant, and I goes up, R would have to go down. But I do not imagine this is true.

7. Feb 20, 2013

### CWatters

No that's wrong. The internal resistance depends on a lot of factors including the battery chemistry but generally the larger the battery the lower the internal resistance. It's not allways true.

With some types of cell the manufacturer is able to trade off capacity vs internal resistance. For example thick plates have lower resistance but take up more space in the cell and so they reduce the capacity. Thin plates have higher resistance but you can get a greater area in the cell.

The expression "more current" can be confusing...

1) A D cell typically has greater capacity (more AH) so if you need the cell to last 1 hour (or any fixed time) then you can draw "more current" from a D than you can from an AA.

2) A D cell probably has lower internal resistance. This means that if you need the cell to deliver at a minimum voltage (say 1.4V) then you can draw "more current" from a D cell than you can an AA cell.

3) If the load has a fixed resistance then a D cell may push "more current" through it than an AA because it's output voltage is higher due to 2).

If you are planning to experiment I would stay away from using bulbs as loads. Bulbs appear simple but the resistance of the filament isn't constant, it can vary with temperature. If you are trying to understand V=IR then it helps if R is constant! If R varies when you change V or I then it's very easy to get confused. Use resistors instead.

8. Feb 20, 2013

### Integral

Staff Emeritus
Were both batteries new? or Used? Did you actually MEASURE the voltage or are you just guessing. In your circuit Ohms law will hold. If the bulb is brighter it is because the you have applied a higher voltage.

The D cell will not provide more current to the circuit it will just maintain the current for a longer time. That is, it has a higher Amp-Hour rating then the AA battery.

9. Feb 20, 2013

### CWatters

Did you actually measure the voltage?

10. Feb 20, 2013

### kjamha

Thank you everyone! I went back and measured the voltage and the AA did not have 1.5 V, so I switched it out and the bulb brightness went significantly up. Both AA and D looked to be about the same brightness. Integral, it makes sense that a D battery is built to simply last longer. I went to ehow to look at the differences between D and AA batteries and drew the wrong conclusion based on what was written about the power of the batteries (D batteries supposedly provide approximately 8 times more power). with larger Power and a similar voltage, I thought the current went up.
Does the current actually go up in a D vs. AA battery? How is does the D have so much more power?

11. Feb 20, 2013

### sophiecentaur

First - Batteries:There are several factors involved with batteries. You can get long lasting ones, ones that will supply high current, cor instance. There are minute hearing aid batteries and watch batteries that only store a small amount energy but they will supply a tiny power for a couple of weeks to a couple of years or more.
Most devices that use D cells are high power - requiring a lot of current and you will appreciate that a D cell has a bigger internal area than AA for the chemical reaction to work over, giving a higher current before the voltage 'sags' (described in terms of an internal resistance in the battery).
It's very hard to make meaningful comparisons between different makes and sizes of batteries.
Next - Resistance: There aren't really any good 'descriptions' of resistance because it's not 'like' anything else. The only sure fire description is that it is the Ratio of Voltage across something to the Current flowing through it.
R=V/I
R doesn't stay the same, even for metals, if the temperature changes and people tend to use the expression "Ohm's Law" out of context. (R=V/I is NOT Ohm's Law, for instance).
Using words like "hinderance" can get you into the wrong way of thinking about this quantity. People will liken it to Friction and the effects of viscosity on fluid flow in pipes. There are similarities, of course, but, as with all matters electrical, it really is more fruitful to 'go with the Maths' as much as you can and don't try too many mechanistic pictures in your head. It's very risky to take on board someone else's pictorial views of electricity - and even worse to pass them on because it can turn into 'Chinese Whispers' and the understanding suffers.

12. Feb 20, 2013

### Crazymechanic

If you say that the two batteries with different capacities had the same voltage but when connecting to a bulb one was brighter and the other one was dimmer then it is almoust certain that you measured the voltage on them with no load if you did it at all.
Smaller batteries usually can have the same voltage but not the same amperage and when measuring them with no load the voltage can show up as almoust the maximum rated for the battery but when connecting a load it sags dramatically.
This is especially true for older laptops as I have repaired them and the battery is usually the worst part as you can charge it up it shows ok volts but when under load it discharges fast and doesn't last long, maybe your battery was or is old or worn.
Anyway I don't think there is something wrong with the math here or physics too simple of a thing to mess up.

13. Feb 20, 2013

### Naty1

" it really is more fruitful to 'go with the Maths' as much as you can.." in fact, you MUST....

The only way to answer the first question is to get the specifications for D versus AA internal resistance and use the models [calculations] I provided in post # 5. The math is easy....and describes the relative power dissipation in the battery versus in the load.

The answer depends on the maximum power transfer theorem...how closely the load resistance is matched to the battery resistance. Did you read those links? Do you understand?? That's why I previously posted READ CAREFULLY. Once you understand, it's really cool!! [If you know differentiation, you can derive the power transfer theorem yourself. I still remember doing that waay back in college.]

Different sized loads can produce different answers for the same two batteries!

POWER is energy delivered per unit time....electrons delivered per unit time to your load....a bigger battery has more chemical constitutents and so can provide more electrons.....internal to the battery, more chemical in a larger volume and cross sectional area means more ions are available from the electrolyte...so the shape of the battery will also affect how much POWER can be delivered....

Last edited: Feb 20, 2013
14. Feb 21, 2013

### kjamha

I believe my error comes from interpreting an increase in current (D vs AA battery) from P=IV. With the increase in power from the D battery, and V being a constant, I see an increase in I. And with V=IR, once again V is constant, and I increases, therefore R must vary (it must decrease). But I did not look at the resistance of the entire circuit (the internal resistance to be more specific). The internal resistance of the larger battery does go down, allowing more current to flow. I think that partially solves my mystery, but there is more to this puzzle which involves the relative power dissipation in the battery versus in the load. I am obviously a bit green regarding circuits, so I am still trying to wrap my head around power dissipation. Is my summary somewhat accurate? I'm hoping I am not missing anything major. I think ultimately, the D Battery can supply more total energy than a AA can. I know it might be inaccurate to try to visualize what is happening, but I guess I look at this as a AA providing 1.5 J of energy to a small amount of charge and a D battery providing 1.5 J of energy to a larger amount of charge.

15. Feb 21, 2013

### sophiecentaur

Very few electrical energy sources are operated under "maximum Power Transfer" conditions. (Can you name a single one?) This is a really inefficient way of doing things as it wastes as much power as you are delivering. The only 'matching' that is important is to make sure that the internal resistance of the supply / battery is significantly lower than that of the load, which ensures that the PD is acceptable abd that you waste as little power as possible. You shouldn't introduce these things as they just serve to confuse.

Power is definitely not number of electrons delivered. "Number of electrons delivered per unit time" is rate of charge tgransfer - which is Current!! If you want to be helpful, you must be accurate.

16. Feb 21, 2013

### Naty1

If the internal resistance of the larger battery is smaller, fine; That would be my guess, but I don't know for sure. Just be sure you are comparing batteries of the same materials.

I'm not sure what you mean by 'more to this puzzle'....I'd say if you know the resistances [internal battery and load] and you know the current flow, you know everything...you can calculate the power dissipation of each resistance from P = i2R.

But the power transfer theorem suggests that depending on your load size, there could be some limits to the delivered power.....I've never studied such details or if I did I have forgotten.

You seem to be on the right track.

If you want to refine your practical knowledge, check this out:

at http://en.wikipedia.org/wiki/Internal_resistance

Here is how I would try that:

What this is talking about is that when you read open circuit battery terminal voltage, nothing connected except your voltmeter, you might get 1.53 volts as an example. (I usually find new batteries are a bit above 1.500 volts.] But when that battery is connected to a load and delivers current via a circuit maybe you now read...1.48 volts....That lower figure reflects the voltage loss due to the internal resistance inside the battery. Assume you measure the current fed to a load is, say, 100 ma...that is, 0.1 amp. Then the internal resistance of the battery is V/I = [1.53-1.48]/100 ma.... = 0.5 ohms.

I just made up those numbers, but Wiki in the above article says [for an AA battery] :

so 0.5 ohms I produced above doesn't seem a crazy out of bounds figure....

17. Feb 21, 2013

### Naty1

yes, we agree....What I was cautioning the OP, who seemed ready to make some circuit test readings, is that there are some practical limitations...

What I posted is correct; what you quoted is not what I posted..

But rereading my earlier post I could have explained 'current delivered to a load RESISTANCE' which is closer to what I meant, that is a good suggestion.

18. Feb 21, 2013

### sophiecentaur

@Naty1

You clearly know the facts. It is also important to make sure how you present them is helpful - that's why I picked you up.
It may have not have been what you meant to write but my quote was straight off your post. The row of dots between the two descriptions implies that they are equivalent and you know they are not.

Also, the Maximum Power Theorem is really not relevant here. All that counts is the internal resistance and just piling on the extra information about the MPT implies that is how batteries are operated and is confusing.

When one is trying to 'explain' things, it's walking on eggshells. Someone who is uninformed can pick up the wrong end of the stick so easily and these are two examples. I hate it when that happens.