Electricity basis resistors/power/current question

[Femme_physics]

Homework Statement

A voltage of a car battery is 12V, the current requires for movement is 200A. When we turn the lights on the voltage drops to 11.1V. The power of each backlight is 55 watt, the power of each front light is 12 watt.

A) Calculate the internal resistance of the battery.
B) What is the max current that the battery can supply?

The Attempt at a Solution

I'm trying to understand first of all how can voltage DECREASE when you turn the battery on. Isn't the potential difference a constant thing between a certain + and a certain -?

I at any rate tried to solve it, but I have an inkling I'm not quite there.

 

[I like Serena]

Hi FP,

I see you're taking electronics to the next level.
Let's see if I can explain this.

Before I start, however, are you sure about the numbers in the problem?
It seem rather odd that a back light would dissipate so much more power than a front light.
Perhaps the numbers should be reversed? That would be close to real life.
That is 55 W for each front light, and 12 W for each back light.

I'm trying to understand first of all how can voltage DECREASE when you turn the battery on. Isn't the potential difference a constant thing between a certain + and a certain -?

Until now we took batteries for granted and said they would give a constant voltage.
However, in reality that only holds true when the current is sufficiently small.
When the current is significant, the voltage of the battery drops (from 12 V to 11.1 V in your problem, so that's 0.9 V).

This behavior of batteries is modeled by saying they have a so called internal resistance.

So it's just as if we have a perfect battery that really delivers 12 V, that is in series with an internal resistor, which is then connected to the lights.

I at any rate tried to solve it, but I have an inkling I'm not quite there.

If you can find the current flowing, you can use Ohm's law (V = I x R) to determine the internal resistance.

Note that there is however a 2nd real-life issue.
The lights are specified with watts instead of ohms.
The reason is that their resistance isn't quite constant, so they are modeled as if the power they dissipate is constant, which is closer to what really happens.

So the lights get 11.1 V and dissipate 134 W.
With the power formula (P = V x I) you can calculate the current that will flow (without the engine), and use that to find the internal resistance (as I wrote earlier).

 

[Femme_physics]

Hi FP,

I see you're taking electronics to the next level.

Actually, this is the last question in my electronic HW papers! I guess I should be able to solve it, or maybe our teacher is tackling us?..

I've also reached the end of mechanics notebook. The only questions left there are "high-level" questions with shear force included that I'm now trying to work out.

That is 55 W for each front light, and 12 W for each back light.

Nope, the question says 55 to the back, 12 to the front

Until now we took batteries for granted and said they would give a constant voltage.
However, in reality that only holds true when the current is sufficiently small.
When the current is significant, the voltage of the battery drops (from 12 V to 11.1 V in your problem, so that's 0.9 V).

This behavior of batteries is modeled by saying they have a so called internal resistance.

So it's just as if we have a perfect battery that really delivers 12 V, that is in series with an internal resistor, which is then connected to the lights.

Ohhh...so when we say 12V we're not actually considering internal resistance? Do "perfect batteries exist"?

If you can find the current flowing, you can use Ohm's law (V = I x R) to determine the internal resistance.

Note that there is however a 2nd real-life issue.
The lights are specified with watts instead of ohms.
The reason is that their resistance isn't quite constant, so they are modeled as if the power they dissipate is constant, which is closer to what really happens.

So the lights get 11.1 V and dissipate 134 W.
With the power formula (P = V x I) you can calculate the current that will flow (without the engine), and use that to find the internal resistance (as I wrote earlier).

That seems real simple. I just used 12 instead of 11.1, not sure what's the difference. I think I understand now with the earlier comment.

 

[I like Serena]

Ohhh...so when we say 12V we're not actually considering internal resistance? Do "perfect batteries exist"?

Only in theory :)
Oh, and when there is no current flowing (if there is no load connected), the battery *will* deliver 12 V. But then you're not doing anything with it. ;)

That seems real simple. I just used 12 instead of 11.1, not sure what's the difference. I think I understand now with the earlier comment.

Can you find the internal resistance then?
Because in your scan you calculated the total resistance instead of the internal resistance.

The next step would be the second part of your problem, which is the max current the battery can supply.

 

[Femme_physics]

Only in theory :)
Oh, and when there is no current flowing (if there is no load connected), the battery *will* deliver 12 V. But then you're not doing anything with it. ;)

I see, so the moment there's any sort of resistor in the shindig, the potential drops. Right, that's ohm's law come to think of it. *slaps forehead*

Can you find the internal resistance then?
Because in your scan you calculated the total resistance instead of the internal resistance.

If my gut feeling is correct that would have to be:

Internal resistance 12-11.1 = 0.9
And so,

rw = 0.9/134

will give us internal resistance. Yes?

 

[I like Serena]

If my gut feeling is correct that would have to be:

Internal resistance 12-11.1 = 0.9
And so,

rw = 0.9/134

will give us internal resistance. Yes?

I'm not sure what you did here?

You seem to have combined the voltage across the internal resistor with the power dissipated in the lights.

I've attached this image to try to clear it up:

You can combine the voltage across the lights (11.1 V) with the power dissipated in the lights (134 W) to yield a current I through the lights.

This same current I flows through the internal resistor which has a voltage of 0.9 V across, which can yield the resistance of the internal resistor.

 

[Femme_physics]

I'm not sure what you did here?

I was trying to use the power formula with V and R, but I forgot to square V.

You can combine the voltage across the lights (11.1 V) with the power dissipated in the lights (134 W) to yield a current I through the lights.

Seems simple enough. I'm off the scanner so I'll just type it here.

So,

P = IV
134 = I x 11.1
134/11.1 = I
I = 12.072 [A]

Speaking of "power dissipation", is it directly related to energy conversion efficiency? Let's say you have to invest 1100 [W] to operate something that requires 1000 [W]. 100 is power dissipation, yes?

This same current I flows through the internal resistor which has a voltage of 0.9 V across, which can yield the resistance of the internal resistor.

R = 0.9/12.072 ohms


So the voltage that drops during the turning-on of the car, that is to say delta V, is in fact the potential difference in the internal resistance?

By the way, "internal resistance" is basically/mostly the type of wire being used, I understand?

 

[I like Serena]

Seems simple enough. I'm off the scanner so I'll just type it here.

So,

P = IV
134 = I x 11.1
134/11.1 = I
I = 12.072 [A]

R = 0.9/12.072 ohms

Yep. That's it :)

Speaking of "power dissipation", is it directly related to energy conversion efficiency? Let's say you have to invest 1100 [W] to operate something that requires 1000 [W]. 100 is power dissipation, yes?

It is directly related to energy conversion, although not (necessarily) to efficiency.
There's also a language issue here, I think, about whether power dissipation would always be in the form of excess heat, or if it is just any energy conversion. I'm not entirely sure about those word meanings.

The way I would put it in your example, I'd say that 1100 W of electrical energy would be dissipated in the engine (from the point of view of the electrical circuit).
1000 W would be dissipated in the form of movement energy, and 100 W would be dissipated as excess heat.

So the voltage that drops during the turning-on of the car, that is to say delta V, is in fact the potential difference in the internal resistance?

Yes! :)
(That is, I'd say that the "voltage drop" is the potential difference "across" the internal resistance, but perhaps I'm splitting hairs here.)

By the way, "internal resistance" is basically/mostly the type of wire being used, I understand?

I'm not entirely sure what you're asking here.

The "internal resistance" of a battery is an abstract concept that models how a battery behaves, which will be dependent on how the battery is constructed internally.

If you're talking about the wire from the battery to the lights, that will have a resistance as well, which you might also term the "internal resistance" of the wire, but I think that resistance is negligible in this problem.

And anyway, if you look at the specifications of a battery, or indeed of any electric device, it will (usually) specify the internal resistance or internal impedance.

 

[Femme_physics]

The way I would put it in your example, I'd say that 1100 W of electrical energy would be dissipated in the engine (from the point of view of the electrical circuit).
1000 W would be dissipated in the form of movement energy, and 100 W would be dissipated as excess heat.

So it sounds like "power dissipation" is in fact same as "invested power".

Yes! :)
(That is, I'd say that the "voltage drop" is the potential difference "across" the internal resistance, but perhaps I'm splitting hairs here.)

I'm all in favor of splitting hairs if you got the talent to do it :)

'm not entirely sure what you're asking here.

The "internal resistance" of a battery is an abstract concept that models how a battery behaves, which will be dependent on how the battery is constructed internally.

If you're talking about the wire from the battery to the lights, that will have a resistance as well, which you might also term the "internal resistance" of the wire, but I think that resistance is negligible in this problem.

And anyway, if you look at the specifications of a battery, or indeed of any electric device, it will (usually) specify the internal resistance or internal impedance.

I see, so internal resistance is a mixture of a lot of things. Dooly noted.

Real thanks for the help over all of that! :)

 

[I like Serena]

So it sounds like "power dissipation" is in fact same as "invested power".

I believe so, but any time a PF Mentor or an English teacher might swoop into tell me I'm wrong. :)
But whether true or not on a language basis, it's the right way to think about it.


That leaves the second part of your problem, what the maximum current is that the battery can supply.
This would be the case if the lights are switched off and the battery is short circuited.

Since you now know the internal resistance you should now be able to calculate the maximum current...

 

[Femme_physics]

This would be the case if the lights are switched of and the battery is short circuited.

So when we're asked for "maximum" current we treat a circuit as though it's turned off?

P = I^2 x R
I^2 = P/R
I = SQUARE ROOT OF P/R
I = SQUARE ROOT OF 134 OVER 0.9/12.072

Wait actually... you've said this is with the lights being turned off and the battery short circuit...so I don't have P... I do have V when the thing is off, so

Imax = V/R
Imax = 12 OVER 0.9/12.072

Is that right?

 

[I like Serena]

So when we're asked for "maximum" current we treat a circuit as though it's turned off?

We're asked for "the maximum current that the battery can supply".
So we remove the battery from the circuit and prod it to see how much current it will give.
It will give it's maximum current when it is short circuited (and then probably explode or burn out ).

P = I^2 x R
I^2 = P/R
I = SQUARE ROOT OF P/R
I = SQUARE ROOT OF 134 OVER 0.9/12.072

Wait actually... you've said this is with the lights being turned off and the battery short circuit...so I don't have P... I do have V when the thing is off, so

As you already remarked, you don't have a P that is dissipated in the lights. :)

Imax = V/R
Imax = 12 OVER 0.9/12.072

Is that right?

Yep! :)
Careful how you calculate that though, you need to keep your parentheses straight (or "round" I guess ).

What are your final, calculated values for internal resistance and maximum current?

 

[Femme_physics]

We're asked for "the maximum current that the battery can supply".
So we remove the battery from the circuit and prod it to see how much current it will give.
It will give it's maximum current when it is short circuited (and then probably explode or burn out ).

That's interesting!

What are your final, calculated values for internal resistance and maximum current?

Internal Resistance = 74.44 milliohms
Imax = 160.96 [A]

 

[I like Serena]

That's interesting!


Internal Resistance = 74.44 milliohms
Imax = 160.96 [A]

Is there a question C that asks whether the car can move (since your problem states that it needs 200 A)?

 

[Femme_physics]

Actually, no! But there should :)This is the last question I my electronics paper for the holiday, we go back to school Friday. I got nothing left to solve for now! I feel...so... empty...so...questionless...so challengeless...need...more questions to solve... :( *feels breathless*

Thank you so much for helping me out with this.