Charging time for a 378 minute battery pack at 100 mA

[Travian]

Homework Statement

How long will it take the primary energy source to fully charge the battery pack from an uncharged state? (State your answer in hours, minutes and seconds, as appropriate.)

t = 378 min.
I = 100 mA

Homework Equations

Cb (battery capacity) = I x t

The Attempt at a Solution

Cb = I x t
Cb = 100 mA x 3600 s. / 378 min.
Cb = 95,23 minutes = 1 hour 35 minutes 23 seconds

Is this the correct answer?

 

[omoplata]

Is t the time the battery was charged (wink, wink)

 

[Travian]

Yes.

This is how i got it:

Cb = 1050mAh
I = 100mA


T = Cb / I
T = 1.05A x 3600s x 0.1A
T = 378 minutes

 

[omoplata]

If they've given you a value for Cb, why did you calculate it again?

Seems to me that all you have to do is convert t to the hours, minutes and seconds format.

 

[Travian]

so how do i convert mAh to time?:/

 

[omoplata]

You have an answer for t in minutes. Convert that to hours, minutes and seconds.

 

[Davide82]

It is not clear what data you have from the beginning and what quantity you're looking for.

If battery capacity is C = 1050 mAh and charging current I = 100 mA you can calculate the time required in hours with: t = C/I = 1050 mAh / 100 mA = 10,5 h.

But I see inconsistent things: you say Cb is capacity (I x t) and you write Cb = 95,23 minutes?

 

[omoplata]

Yes. Your arithmetic is wrong. Sorry, I didn't look at that. I assumed it was correct.

The answer is t = 10.5 hours.

You have to convert it to the hours, minutes, seconds format.

 

[Davide82]

I really don't know what is the correct answer!
I don't understand what are Cb, I, t... what is a "given" and what is the aim.

In the first post you are looking for a time... BUT you write t = 378 minutes from the very beginning... (what is this time? is it the answer and you want to get it? But then it is different from my calculated 10.5 hours).
In the first post it seems Cb is the battery capacity (in mAh I assume) BUT you write Cb = 95,23 minutes!

Then, in your second post, you write Cb = 1050 mAh... where did you get this?

For completeness I say that these are pure ideal data and real batteries differ a lot in behaviour.

 

[Davide82]

By the way, this thread of yours seems related:
https://www.physicsforums.com/showthread.php?t=425341

 

[Travian]

Ok it's confusing. Let's start again.

Here is entire exercise:

A primary energy source is able to provide 0.7 Watts power in order to charge a mobile telephone battery pack. The telephone operates at 3.5 V, and the mobile telephone’s 3.5 V battery pack has a capacity of 1050mAh. In “talk” the mobile phone draws an average current from its battery of 100mA.

(i) How many minutes of talk time will this battery pack provide, at the average current for “talk”?
(3 marks)

(ii) How much energy (in Joule units) is stored in the battery pack, when fully charged?
(4 marks)

(iii) How long will it take the primary energy source to fully charge the battery pack from an uncharged state? (State your answer in hours, minutes and seconds, as appropriate.)
(4 marks)

(iv) By assuming that all the power, which can be delivered with a 5A current from a 240 V a.c. mains supply, is supplied to charge the battery pack, determine the fastest theoretical time for the battery pack to be charged from zero to its capacity.
(4 marks)

I am currently doing number 3.

How can i calculate time? What formula do i use in this situation? Sorry for this confusion.

 

[Davide82]

Power available = 0.7 W.
Since it should deliver energy at 3.5V (because you're charging a 3.5V battery pack; in real situation the voltage needed to charge a battery is slightly more than the nominal voltage of the battery) we get a charging current I = P / V = 0.7 W / 3.5 V = 0.2 A = 200 mA.
You know Cb = I x t. As first approximation we say it is valid both during discharge and charge processes; so t = Cb/I = 1050/200 = 5,25 hours...
Now you only have to translate to minutes and seconds...

 

[Travian]

Oh now i see where i made vital mistake. I just now noticed that I = 100 mA in "talk"...

If the phone is not in "talk" then i have to calculate another I by using I = P / V

 

[Davide82]

Sure! First thing to do: read very thoroughly the problem!

 

[Travian]

Im a bit lost with number 4.

P = I x V = 5A x 240 V = 1200W
(not sure how or where to use this??)

now should i calculate again t=Cb/I?

 

[Davide82]

As usual, the time needed to charge is t = C / I. The more the current the less the time required.
With 1200 W of power available, how much current can you draw at 3.5 V?
P = V x I; so I = P / V (just as in point (iii)) so I = 1200 W / 3.5 V = 343 A.
So t = 1.05 Ah / 343 A = 0.00306 h = 11 seconds.

(And I say: the battery will explode!)

 

[Travian]

but there already is 5A, why do i need to calculate another A?

 

[Davide82]

First of all I must say that I feel like you have a wrong approach at this exercise and maybe in general in your study. It looks like you are trying to put together all the data you have in some manner.

I suggest you to understand much more the underlying principles.

For example, in this exercise, you have to know a very basic principle which states that energy is conserved. This means that all the energy you get from the 240 V power outlet is going into the battery.
Since you are draining 5A at 240V it means you are getting 1200 W.
At 3.5V the same 1200W means 343A of current.
The transformer is transforming the electrical energy in another form: less voltage and more current. BUT THE POWER IS THE SAME!

If only 5A would have gone in the battery, then you were draining 1200 W and only putting 3.5V x 5A = 17.5 W in the battery. Where would be going all the missing power?

The problem was saying:
"By assuming that all the power, which can be delivered with a 5A current from a 240 V a.c. mains supply, is supplied to charge the battery pack."

Going a little bit deeper, we are assuming that the transformer is 100% efficient and good at its job; in reality, in the real world, the transformer heats a little and a part of energy is going wasted into heat, so, for example, maybe only 1100W are effectively reaching the battery... But this is a topic which usually is not considered in these, so simple problems. Maybe you will encounter "efficiency" in the future... here it is...

 

[Davide82]

And... I was a bit imprecise: Watts are a measure of power, not energy; but since energy is conserved and the draining from the outlet is taking the same time as the charge process, you can say that the power is conserved, too.