Calculating Talk Time for a 1050mAh Mobile Phone Battery Pack

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Homework Help Overview

The discussion revolves around calculating the talk time for a mobile phone battery pack with a capacity of 1050mAh, operating at 3.5V, and drawing an average current of 100mA during use. Participants are exploring the relationship between battery capacity, current, and time, as well as energy calculations related to the battery.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula relating battery capacity to current and time, with some attempting to derive time from the given capacity and current. There are questions about the correct application of formulas and units, particularly regarding energy calculations.

Discussion Status

There is an ongoing exploration of the correct formulas to use for calculating talk time and energy stored in the battery. Some participants have provided guidance on the relationships between the variables, while others are questioning the units and conversions necessary for their calculations.

Contextual Notes

Participants are navigating through potential misunderstandings regarding the use of units and the application of formulas, particularly in converting battery capacity and voltage to appropriate SI units for energy calculations.

Travian
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Homework Statement



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”?




Homework Equations


P=0.7
I=100mA
V=3.5V



The Attempt at a Solution


Now I'm not sure what formula to use in order to find time. Can someone help me?
 
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The capacity of the battery is given in mAh. This means units of current (amps) times units of time. So what would be the formula that defines the capacity of the battery?
 


This formula.
(Cb) = I x t

then
time = C / I
?
I need to find time
 
Last edited:


Travian said:
This formula.
(Cb) = I x t
Yes, this is it. It is actually electric charge (Q) but in battery industry they call it capacity.


Travian said:
then
time = I / (Cb)
?
I need to find time


No, this is not correct. Watch your algebra.
You need to find time and you can do it from the formula above (Cb=I x t). Solve for t.
 


Sorry i changed it to time=Cb/I

How can i use this formula (Cb=I x t) if i don't know time?
Shouldn't i use this? t=Cb/I (t = 1.05Ah / 0.1A = 10.5)
 
Last edited:


Travian said:
Sorry i changed it to time=Cb/I

How can i use this formula (Cb=I x t) if i don't know time?
Shouldn't i use this? t=Cb/I (t = 1.05Ah / 0.1A = 10.5)
Yes, now it's OK. You use Cb=I*t to calculate the time, t= Cb/I, as you wrote above.
First time you wrote t=I/Cb which is not correct.
 


so the answer i suppose is 10.5 hour = 650 minutes.
 
Last edited:


There are more to this exercise.

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

I used this formula:

E = Cb x V
E = 1.05Ah x 3.5V
E = 3.675 (what units?)
 


Travian said:
There are more to this exercise.

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

I used this formula:

E = Cb x V
E = 1.05Ah x 3.5V
E = 3.675 (what units?)

If you convert your inputs in SI units (something you did not do here) then the answer should be in SI units of energy, as required in the question.
 
  • #10


SI units.. Ah to mAh and V to mV?
 

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