V=IR and Power are these statements correct?

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nirVaan

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hi guys im trying to figure out the relationship between power and V=IR because i have seen in some books statements like

"the power needed to produce the current...."

so how i understood this is like

1. in a ideal perfect situation V=IR holds true and a certain V can produce any amount of current depending on the resistance.

2. but in practical situation there is a limit to the current that can be produced and that depend on the power that can be provided by the source.

3. different sources with the same voltage (ex. 1 volt) may have have different amounts of ability to provide power in them

4. which means even thought both have same voltage the maximum current it can generate will vary.

5. so when designing a circuit you 1st figure out the total current needed by the circuit and the total resistance and then figure out how much voltage is needed.

6.then you figure out the total power (w=VI) and find a voltage source of the desired voltage that can also produce this much of power.

7. and also you make sure the resisters that are used can with stand this power with out getting heated and melted.

thats what i can think of i do not know whether this is correct (at least at a higher level of abstraction) so can some with experience and knowledge tell me whether this is correct or incorrect and if it is incorrect why is that so ?

Thankz
 
A dinky little 12V battery is made for the remote control for a garage door opener. It can supply 10mA for about 1 minute. Its current is usually a few mA for 1 second and is used once per day so it lasts for months.

A car battery is also 12V. A big one can supply 100A or more for one hour.

Look at the huge difference in capacity.
Look at the huge difference in size and weight.
Look at the huge difference in cost.
 

Have a look at the site that Blueroomelectronics linked in his post

Robert-Jan
 
thankz for your reply.

one final point i like to clarify though

when voltage source says it has 5V does this mean that if the voltage source is connected to a resistance of one ohm it is guaranteed to provide a current of 5 A regardless of its internal resistance.
 

hi,
No.
You must define the characteristics of the voltage source.

EDIT:
All voltage sources have an internal resistance.
Any current flowing thru an external load will also flow thru this internal resistance.
The current in Rint will create a voltage drop across Rint, so the 'output' voltage will fall.
example:
if the Rint = 1hm: and the external load = 1hm: and the Vsrc was say 5V,
then using Ohms law formula, I = 5/[1+1] = 2.5Amps.

So the voltage across the external load would 5V - 2.5v = 2.5V

Do you follow.?
 
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yes i do but why cant the voltage source specify the voltage taking in to account the internal resistance lets say a battery has a internal resistance of 1 ohm and with out any load its voltage is 10 V . but you specify it in the battery cover as 5 v so some one new to electronics would not have to be concerned about internal resistance etc... he sees that the voltage is 5 volts in the cover of the battery he just connect a 1 ohm resister to it.and he gets a current of 5 A just as he expects ( because the batteries actual voltage is 10 V and has a internal resistance of 1 ohm). specifying the voltage like this in a voltage source would not have any adverse effect in a circuit as well or would it ??????
 
the voltage drop that fals over the internal resistance is depending on how many current is drawn from the source V=I*R

if you would specify your battery as 5V with no other info but actualy its 10V open poles voltage

what would happen if your resistance is 100 ohm??

the voltage would be so high that it could blow out TTL IC's

someone new to electronics should learn these things very intensive to understand this matter in my opinion it's importand to know these basic prinsipels

its not the most funky part of electronics but if i could give one advice to learn this it will help you later much quicker through other theorys

kirchof and thevenins laws are also not that funky but again very importand to understand much complexer things in the electronics

Robert-Jan
 
this is interesting. few questions come to mind.

1.is the voltage difference the same thats falls over the internal resistance as well as the external resistance when a external circuit is connected?

2.so does the voltage of a voltage source i.e battery changers according to the current its suppling?

3.if so what does the voltage value given in a battery (eg: 1.5 V)really signifies ,what does it tell a person and what can we understand from that given value ?


and when it comes to power do we use W=VI to find the power that can be generated by a voltage source and use w=(I^2)R to find the power dissipated external circuit components. and the maximum power generated by the source should be more than the total power dissipated by the external circuit.
 
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hi,
If you are talking about batteries, you have to consider the type of battery, that is, what active materials are used in its construction.
eg: NiCadm, ZiFe, lead acid.
They all have different internal resistance values, which change with the batteries state of charge and its age.

If its a power supply, these are rated for a given current they are able to maintain a stated voltage.
 

you can use both depending on which parameters are know by you at the time of the calculation

in the case of a battery you should also take time in consideration
a battery is stored power = energy
energy is power during a given time

energy is the posible potential (example a 100 Ah battery can deliver 100 Amps for the duration of 1 hour but it can also deliver 10 Amp over 10 hours in both calculation the energy is the same)

Robert-Jan
 
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so is it ok if i say

the amount given in a voltage source eg 1.5 v says that the voltage on the two poles will be 1.5 V plus or minus some value. and is able to provide a current of nearly 1.5A if a external resistance of (1-internal resistance) ohms are connected to it provided that the voltage source have enough power to produce a current of 1.5A for a given time period. (1 minute ,1 hour etc....) ?


but i still cant understand why there is a voltage drop i thought for a given voltage it will stay the same and current is the one that will change according to the resistance
 
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the current that flows is calculated over the total resistance that the voltage source sees

that means internal resistance and external resistance because you have 2 resistors means that over each of them will fall a voltage depending on the current that flows through it if your talking about the voltage on the poles than your talking over a part of the system (you exclude the internal resistance)

if you talk about a power source you have to see it like the picture shows in eric's post

you see a perfect voltage source (no resistance but this one don't exist) and you see the internal resistance, both together is a power source with poles in real life

if you see the calculation U=I*R

and R stays the same but I change than U will change also

if the current I stays the same but the resistor change than U will vary also

Change of the Resistor value in a power source is posible due to tempature of materials that build the source (see your Phisics book regarding this )

Robert-Jan
 
hey thankz

i think i got it now.all this time i was thinking the internal and the external resistance are in parallel.but in fact they are in series.

so if you have 1 V source and internal resistance of 1ohms and you connect it to a external resistance of 2 ohms.

then the current flowing through it is i=v/r i=1/(1+2)
which is 0.3 A but the voltage across the external resistance is

V=IR V=0.3*2
which is 0.6 V

this is what you mean right. well this makes sense now

but one more small question can you tell me in simple terms how does a internal resistance can limit the maximum current that can be provided by a voltage source. i mean if the internal resistance is 1 ohm and voltage is 1 V the maximum current it can produce is 1 A right. but in books i have seen that they have said the internal resistance is for a 1.5 V battery is about 1 ohm and yet the maximum current it can provide is around 250mA how can this be.
 

I am not 100% sure about this but this is my idea about it
at first a battery is a energy storage so that means there are limits how much it can store and how much it can releas in a certain time

if you have a big drain of energy than we saw already that we would have a voltage drop

but at the same time energy is thermaly disapated in the internal resistance which would have than a tempature rise which would affect the resistance again in increasing the resistance of the internal resistance and so limit the current

the voltage potential of a battery is set by the difference in materials potentials that are used in the battery and the voltage can never drop till a 0V In the power source so there will be always a potentiall differance over the internal resistor

so it's reach a point that due to the matterials that are used the voltage is not able to drob more and that the current can not rise more because of a balance point that is reached that if a bigger current is drawn the resistor get higher in valeu and thus limit the current again thats the max current a power source could give than

for how long??

that depents on the size and sort of storage media of the energy

Robert-Jan
 

Batteries do not store electricity [voltage] they store electro-chemical energy.
When a load is applied to the battery terminals this electro-chemical energy is converted to electrons.
The electrons are the current carriers that flow in the load, they flow thru the load due to the electro potential gradient formed between the plates of the battery.
Its the 'resistance' [disassociation and recombination at molecular level] to the conversion of electro chemical energy to electron flow that is the primary source of the batteries internal resistance.
Remember, a battery cell is essentially two plates, of different materials, separated by a conductive electrolyte.
 
Your books are way out of date.
On Energizer's website they show an obsolete no longer made AAA carbon-zinc battery cell with an internal resistance of 0.5 ohms when new. Their alkaline AAA cell has an internal resistance of only 0.225 ohms.

If a 1.5V cell has an internal resistance of 1 ohms and has a 1 ohm load then the voltage to the load is only 0.75V and its current is only 750mA.
If the cell has a load of 5 ohms then the voltage to the load is only 1.25V and its current is only 250mA.
 
nirVaan said:
so can the alkaline AAA cell produce a current of 1 A?
Click to expand...
Yes, for a very short time.

Energizer's datasheet for their AAA alkaline cell shows a 150mOhm to 300mOhm internal resistance when fresh. So the load resistance is 1.3 ohms and its voltage is 1.3V when the battery cell is brand new due to the voltage divider action. The voltage (and resulting current) will keep dropping to about 1.0V in about 5 minutes.
 
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