# Electrical Power - Equation Questions:

• I-need-help
In summary, a 1.5V rechargeable cell labelled 2300 mA h can supply 2300 mA for one hour. When calculating its energy storage in mWh and J, it is equivalent to 3450 mWh and 12.42 kJ respectively. When connected in series, four of these cells can light a bulb labelled 6.0V and 120 mA for 19.2 hours.

#### I-need-help

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

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!