Electrical Power - Equation Questions:

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SUMMARY

The discussion centers on calculating the energy storage and usage of a 1.5V rechargeable cell with a capacity of 2300 mAh. The energy stored in the cell is determined to be 3450 mWh or 12.42 kJ. Additionally, when four cells are connected in series to power a 6V, 120mA bulb, they can sustain the bulb for approximately 19.2 hours. This calculation is based on the current requirements of the bulb and the total capacity of the cells.

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A 1.5V rechargeable cell is labelled 2300 mA h. This means that it can supply the equivalent of 2300 mA for one hour.

b) How much energy does the cell store in:
i- mWh?
ii- J?

and c) How long can four cells, connected in series, light the bulb?

for b) i- I got: W = V x Q, so W = V x I x t

W = 1.5V x 2.3A x 1second
= 3.45W = 3450mWh​

b) ii-
W = Js-1
J = W x t
J = 3.45 x 60 x 60
J = 12420 = 12.42kJ​


I know that I got b (i) and (ii) correct, expect I'm stuck on c). The previous sub-questions are just supplied to give more information :) Thanks
 
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What are the characteristics of the bulb (its wattage or current requirement)?
 
gneill said:
What are the characteristics of the bulb (its wattage or current requirement)?

Ah yeah, missed out some info. The bulb is labelled 6.0 V and 120mA... no wattage, but you could work that out, so P = 6V x 0.12W = 0.72W
 
You know that each cell will be depleted after one hour supplying 2300 mA. The bulb requires 120 mA. How long will they last at that rate?
 
Each cell will last... 2300mA shared by 120 mA ... so, 19.2 hours. And that's the right answer in the back of the book... wow. Hmm. Thanks! That is right isn't it? The textbook isn't always necessarily correct.
 
Yup. That's right.
 
Ah thanks again gneill!
 

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