Question on Electrical Power

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#1
Hi everyone, I'm studying AS Physics and I found this question a bit difficult. Can anyone give me some pointers - my problem is that I'm unsure as to which equations to use, and which values to put in (I know P= IV, P=V^2/R, P=I^2R and V=IR). Some of this question I've already been able to do, but it's parts c) and d) which have confused me:

"The filament bulb of a torch is labelled 6.0 V 120 mA.
a) How much energy does it dissipate per second in mW when working normally?
P= IV so P = 6(1.2x10^-1) = 0.72 W = 720 mW. This is correct according to the answers from the textbook I got this question from.
A 1.5 V rechargeable cell is labelled 2300 mA h. This means that is can supply the equivalent of 2300 mA for one hour. (For example, it could supply 23 mA for 100 hours).

b) How much energy does the cell store in
i) mWh: 2300 mAh x 1.5 V = 3450 mWh. This is also correct.
ii) J ? 3.45 Wh x 3600 s/h = 12420 J = 12.4 kJ to 3 s.f. I joule = 1 watt-second. This is also correct

Finally

c) How long can four cells, connected in series, light the bulb? - This is what's confusing me. According to the answers the answer is 4.8 h.
d) Suggest why your answer to c) is likely to be an overestimate.
Because some electrical energy is transformed into heat energy in heating the connecting wires is my guess.
0
7 years ago
#2
(Original post by Mr Cosine)
ii) J ? 3.45 Wh x 3600 s/h = 12420 J = 12.4 kJ to 3 s.f. I joule = 1 watt-second. This is also correct

Finally

c) How long can four cells, connected in series, light the bulb? - This is what's confusing me. According to the answers the answer is 4.8 h.
d) Suggest why your answer to c) is likely to be an overestimate.
Because some electrical energy is transformed into heat energy in heating the connecting wires is my guess.
Convert the battery total capacity and lamp power rating to the same energy units (Joules).

Spoiler:
Show
Total energy capacity of the battery = 12420 Joules.

Power = Watts = Joules/second.

Lamp output power = 720mW = 0.72 Joules/second

At this consumption rate, the lamp will last:

12420/0.72 = 17250 seconds

17250/3600 = 4.8hrs.

d) Your answer is fine. Also: Ah rating is a manufacturers approximation and will vary between batches, shelf life of battery causes capacity to diminish with age etc.
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#3
(Original post by uberteknik)
Convert the battery total capacity and lamp power rating to the same energy units (Joules).

Spoiler:
Show
Total energy capacity of the battery = 12420 Joules.

Power = Watts = Joules/second.

Lamp output power = 720mW = 0.72 Joules/second

At this consumption rate, the lamp will last:

12420/0.72 = 17250 seconds

17250/3600 = 4.8hrs.

d) Your answer is fine. Also: Ah rating is a manufacturers approximation and will vary between batches, shelf life of battery causes capacity to diminish with age etc.
OK, 0.72 W x t = 12400 J => t = 12400/0.72 = 17222.22... seconds
I divide by 3600 s/h to get 4.78... hrs which equals 4.8 hrs to 2 s.f. Thanks for your help!
Also, what exactly do you mean by 'the capacity of a battery'? Do you mean the ability of the battery to supply charge to circuit components (which will reduce as the lithium plates (and other chemicals) fatigue?).

Anyway, you've been a great help. Thank you!
0
7 years ago
#4
(Original post by Mr Cosine)
Also, what exactly do you mean by 'the capacity of a battery'?
Probably a bad word to use because it has connotations of capacitance which is not what I mean.

I am of course referring to the potential chemical energy available from the battery which indeed is a function of the molar mass of the reactants available for electron liberation.

These chemicals continue reacting (at a very slow rate) even when there is no direct conduction path (circuit load) between the electrodes.
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#5
(Original post by uberteknik)
Probably a bad word to use because it has connotations of capacitance which is not what I mean.

I am of course referring to the potential chemical energy available from the battery which indeed is a function of the molar mass of the reactants available for electron liberation.

These chemicals continue reacting (at a very slow rate) even when there is no direct conduction path (circuit load) between the electrodes.
OK, I understood that, thanks for all your help!
0
6 years ago
#6
I'm coming to this a bit late, but just in case anyone finds this thread(!):

(Original post by uberteknik)

Total energy capacity of the battery = 12420 Joules.
Part (c) says "How long can four cells, connected in series, light the bulb?"

Doesn't that mean that the total energy capacity of the battery, in this scenario, is 4 * 12,420 J = 49,680 J ?

If so, that would imply:

Spoiler:
Show

49,680 J / 0.72 Js^-1 = 69,000 s

69,000 s / 3,600 s = 19.1(6) hrs

= 19.2 hrs (3 s.f.)

I could be wrong, but hopefully someone can provide another opinion on this solution.

The answer in the book is indeed 4.8h, but note that this is 1/4 of 19.2h -- which would be the case if only one cell's worth of energy was used in the calculation. (And, of course, the book can sometimes be wrong! :)

Thanks for any help.
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