Recent content by Travian

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

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?

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

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”...

so how do i convert mAh to time?:/

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

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...

I figured it out. Thank you for your help. I'll keep in mind that Ah = A x 3600s :]

Oh right. E = 7.2V x 2A x 3600s = 7.2V x 7200Ah = 51840 J However what bothers me is... It was 2 Ah before. After i multiply 2 Ah by 3600s, is it again Ah?

Homework Statement Calculate the total energy stored in the BP-927 battery when it is fully charged. P = 9.5 W, V = 7.2 V, I = 1.3194 A Homework Equations E = V x Ah The Attempt at a Solution E = 7.2V x 2Ah = 14.4 Joules..?

Thank you for your help! There is more to this exercise, should i post it here, or create a new thread?

2000 mA x 3600 / 1319 mA = 5458,68s = 90,978 min.

mm so it's 2000/1,319 = 1516s = 25 min?

If this is correct then what does 200mAh have to do with this?

Current: I = P / V I = 9.5W / 7.2V = 1.319A Current x time: Cb = I x t Cb = 1.319A x 3600s = 4748.4s = 79.14minutes