## Main Question or Discussion Point

V/R=I Simple Ohms law right?

This is what I don't get I've been searching everywhere & I still can not find the answer!

If I were to have a 12 volt battery at 5 amps & if I were to apply a 1 ohms resistor to it.

12v/1ohms = 12 amps???!!!

So by adding a 1ohm resistor to my battery I magically increased my amps??!

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berkeman
Mentor

v*r=i simple ohms law right?

This is what i don't get i've been searching everywhere & i still can not find the answer!

If i were to have a 12 volt battery at 5 amps & if i were to apply a 1 ohms resistor to it.

12v * 1ohms = 12 amps???!!!

So by adding a 1ohm resistor to my battery i magically increased my amps??!

Check your equation: V = I * R

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Check your equation: V = I * R
I meant to say v/r=I still would equal 12 amps.

Battery only specified in voltage. They specify amp hour as the energy storage in the battery. They do not specify "at 5A". You miss read that.

Battery only specified in voltage. They specify amp hour as the energy storage in the battery. They do not specify "at 5A". You miss read that.
No actually my battery states very clearly 2,600 mili amps per hour. Which equals 2.6 amps.

Another question is doesn't adding a resistor decrease both my volt & amps of my battery? & How can I calculate that?

berkeman
Mentor

No actually my battery states very clearly 2,600 mili amps per hour. Which equals 2.6 amps.
It's just a way of specifying the energy available. It still comes down to amp*hours. You will get somewhat less available energy at high current drain rates, and somewhat more available energy at low drain rates.

berkeman
Mentor

Another question is doesn't adding a resistor decrease both my volt & amps of my battery? & How can I calculate that?
Adding a resistor where? V = I * R, so the output current is determined by the battery voltage and the resistance across the battery.

It's just a way of specifying the energy available. It still comes down to amp*hours. You will get somewhat less available energy at high current drain rates, and somewhat more available energy at low drain rates.
I hooked up 3, 2v 8amp batteries in parallel & when I tested my amperage I got 24 amps at 2volts. I created something that required a very high current at 16 amps & a low voltage at 2volts. The Amperage listed on batteries is very precise & matches what I get in amps per hour.

For instance when I hooked up a 8 amp 2volt battery to my array of capacitors it took me 30 seconds to fully charge them. When I hooked them up to a 24 amp 2volt battery I was able to charge my capacitors in 2 seconds.

Thus the more amps provided per hour the larger the capacity of a battery.
I am sure you already know this but still does not answer any of my questions.

cmb

No actually my battery states very clearly 2,600 mili amps per hour. Which equals 2.6 amps.

So... err.!! If I drive at 26 miles an hour away from my house, then I must be 26 miles from my house?

Adding a resistor where? V = I * R, so the output current is determined by the battery voltage and the resistance across the battery.
If I were to add a 1ohm resistor on the positive line/wire that is hooked up to a battery, then test for amps & voltage.

berkeman
Mentor

I hooked up 3, 2v 8amp batteries in parallel & when I tested my amperage I got 24 amps at 2volts. I created something that required a very high current at 16 amps & a low voltage at 2volts. The Amperage listed on batteries is very precise & matches what I get in amps per hour.

For instance when I hooked up a 8 amp 2volt battery to my array of capacitors it took me 30 seconds to fully charge them. When I hooked them up to a 24 amp 2volt battery I was able to charge my capacitors in 2 seconds.

Thus the more amps provided per hour the larger the capacity of a battery.
I am sure you already know this but still does not answer any of my questions.
Sorry, none of your math works, and it doesn't match how batteries and capacitors work.

You said your 2V "8A" battery charged the caps in 30 seconds, and your 2V "24A" battery charged them in 2 seconds. 24/8 does not equal 30/2.

Please look at the datasheet for your battery (you can usually find it online at the battery manufacturer's website). It will explain what the Amp*Hour capacity means, and how it varies with discharge rate.

Then always use V = I*R in your calculations. With capacitor charging, the equation is different.

berkeman
Mentor

If I were to add a 1ohm resistor on the positive line/wire that is hooked up to a battery, then test for amps & voltage.
It would be better to use a smaller resistor, more like 1/10 Ohm. Working at higher currents like you are, the errors introduced in the measurements will be smaller.

Sorry, none of your math works, and it doesn't match how batteries and capacitors work.

You said your 2V "8A" battery charged the caps in 30 seconds, and your 2V "24A" battery charged them in 2 seconds. 24/8 does not equal 30/2.

Please look at the datasheet for your battery (you can usually find it online at the battery manufacturer's website). It will explain what the Amp*Hour capacity means, and how it varies with discharge rate.

Then always use V = I*R in your calculations. With capacitor charging, the equation is different.
I never timed it I am just knew it was way way faster. This still does not answer my original question.

berkeman
Mentor

I never timed it I am just knew it was way way faster. This still does not answer my original question.
Could you please re-state your question, now that you know V=I*R? Thanks.

Could you please re-state your question, now that you know V=I*R? Thanks.
I meant to say v/r=I still would equal 12 amps.

berkeman
Mentor

I meant to say v/r=I still would equal 12 amps.
12V / 1 Ohm = 12A correct.

12V / 2 Ohms = 6A correct.

and so on... The higher the resistance put across a voltage source, the lower the current that flows through that resistance.

12V / 1 Ohm = 12A correct.

12V / 2 Ohms = 6A correct.

and so on... The higher the resistance put across a voltage source, the lower the current that flows through that resistance.
Ok what if my battery was only producing 6A then I apply a 1ohm resistor to it will it increase my Amp to 12?

berkeman
Mentor

Ok what if my battery was only producing 6A then I apply a 1ohm resistor to it will it increase my Amp to 12?
Please stop that. As mentioned already in the thread, a battery is a voltage source. It is not a current source. The fundamental thing produced by a battery is voltage. So when it us used in a circuit, the output voltage is the *cause*, and the current that flows as an *effect* is determined by the load resistance (or impedance) connected to the battery terminals.

Please endeavor to understand the cause-effect relationship when dealing with voltage sources like batteries.

No actually my battery states very clearly 2,600 mili amps per hour. Which equals 2.6 amps.
This means if you draw 2.6A, the battery should last an hour.

You are confuse between the battery storage capacity to the current. This amp hour give you the energy stored inside the battery. That has nothing to do with how many amp you can get out of it.

For example a 12V carbon zinc battery, Lead Acid batteries of the SAME amp hour rating of 1 amp hour. If you put a resistor of 12 Ω across, you get 1A through on each and they all last one hour.

But the two are very different. Lead acid has higher internal resistance and even if you short it, it can only give out ( just for example) 5A max. But the Lead acid is known to have much lower internal resistance, it can give out 200A. So if you put a 1Ω across them, the lead acid battery can give out 12A and it only last 1/12 hour. The carbon zinc battery can only give out 5A, so the voltage across the resistor will go down to 5 volts. This mean the output of the battery will drop down to 5V. The 7V that disappeared is dropped inside the battery's internal resistance. BUT it will last longer than the lead acid battery with the 1Ω.

I hope this clarify for you. You are mixing up the energy storage capability of the battery with the peak output current.

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phinds
Gold Member
2019 Award

You are misunderstanding what the battery specs mean. 12V and 5A means that it is a 12V battery that can deliver as much as 5amps. If you put a 12ohm load on it, it will deliver 1 amp. If you put a 5/12 ohm load on it, it will deliver 5 amps. If you put a load on it that requires more than 5amps then various things could happen including (1) the batter voltage goes down and (2) wires start to melt and various interesting fizzing and/or popping noises come from the battery and then shortly thereafter it doesn't put out any amps at all.

danaman

A12 volt battery, 1 ohm resistor will be 12 amps as you say.
It will not increase any battery amps.
It may drain your battery pretty fast..

Others have already stated what the markings on the battery mean.

russ_watters
Mentor

No actually my battery states very clearly 2,600 mili amps per hour. Which equals 2.6 amps.
No, it doesn't. It says miliamp-hours

You're misunderstanding/being sloppy with the units. Amps and Amp-hours are not the same thing -- and there is no such thing as amps/hour.

A battery provides current at basically whatever amount the circuit demands. How long it provides that current is determined by the capacity (in amp-hours) of the battery.

cmb

Ok what if my battery was only producing 6A then I apply a 1ohm resistor to it will it increase my Amp to 12?

danaman - answer my question in #10, or do you not understand the significance?

My car says it can do 120mph, yet the other day I did a 180 mile journey out of it! Amazing!

A 2300mAh battery can produce maybe 20A or more (it depends on the battery's internal resistance as to how much current it can acutally deliver). You are misunderstanding the battery data, making all your questions irrelevant and 'not even wrong'.

You are making the same mistake that many questioners to PF make and assume that the max current is supplied all the time.
I assume that somewhere on you battery it says 12 VDC 5 amp.
What your battery is saying that it is capable of supplying up to 5 amps at 12 volts.
If you connect a 12 ohm resistor across the terminals it will draw a current of 1 amp (V=I*R) (note to student I*V = watts what power rating of resistor do you need for this experiment).