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Capacitor question - NEED HELP!!!

Q. A 2.0 mF capacitor, used as the backup for a memory unit, has a potential difference of 5.0 V
across it when fully charged. The capacitor is required to supply a constant current of 1.0 µA
and can be used until the potential difference across it falls by 10%. How long can the
capacitor be used for before it must be recharged?
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Initially I thought of using the decay of charge equation:
http://prntscr.com/aq1v1h
So the initial p.d. is 5V, and 10% of that is 0.5V, then working out the resistance by using R = V/I = 5V/1*10^-6 = 5*10^6 Ohms. Then rearranging the equation for -t = ln(V/Vnought) * RC, this had given me a ridiculously large value. I'm confused on what this equation is used for, as in what is the "-t" in that equation? When am I supposed to use it?

Secondly, I realized that just using t = Q/I, and firstly working out Q by using Q = CV = (2*10^-3 F) * (0.5 V) = 1*10^-3, and finally putting that into Q/I to give me a time, t = 1000 seconds was the correct answer. I'm not sure also in what situations to use that. Thank you in advance if you can help me out!

Reply 1

TSR Jessica

Sorry you've not had any responses about this. :frown:

Are you sure you've posted in the right place? :smile:

Here's a link to our subject forum which should help get you more responses if you post there. :redface:

You can also find the Exam Thread list for A-levels here and GCSE here. :dumbells:

Just quoting in Puddles the Monkey so she can move the thread if needed :h:

Spoiler

Reply 2

Shaun_24

I have a feeling that you have used 0.5 as the value of v in the decay equation. If you read carefully, it says "until the potential difference across it falls by 10%" meaning that the voltage decreases by 10%. This would mean that the capacitor can be used until the voltage is 5 - 0.5= 4.5V.
Hope this helps

Reply 3

uberteknik

Original post by 1017bsquad

Secondly, I realized that just using t = Q/I, and firstly working out Q by using Q = CV = (2*10^-3 F) * (0.5 V) = 1*10^-3, and finally putting that into Q/I to give me a time, t = 1000 seconds was the correct answer. I'm not sure also in what situations to use that. Thank you in advance if you can help me out!

The question needs you to recognise that current is defined as the rate of change of charge.

i = \frac{dq}{dt}

It also says that the capacitor supplies a constant current of

1\mu C

Finally, the capacitor can supply charge until it's p.d. falls by 10%. In other words, it's charge has also fallen by 10%.

Change in charge = Initial charge - Final charge

Q = CV = (2 \rm x 10^{-3}) \ x \ 5 = 10 x 10^{-3}C

(initial charge)

Q = CV = (2 \rm x 10^{-3}) \ x \ 4.5 = 9 x 10^{-3}C

(final charge)

Charge has fallen by 1mC

Because charge is supplied at a constant rate of 1uC per second, the duration this can be sustained before the capacitor needs recharging is:

\frac{1 \rm x 10^{-3}}{1 \rm x 10^{-6}} = 1000 \rm \ seconds

Original post by 1017bsquad

So the initial p.d. is 5V, and 10% of that is 0.5V, then working out the resistance by using R = V/I = 5V/1*10^-6 = 5*10^6 Ohms. Then rearranging the equation for -t = ln(V/Vnought) * RC, this had given me a ridiculously large value. I'm confused on what this equation is used for, as in what is the "-t" in that equation? When am I supposed to use it?

For this question, we cannot use a fixed resistance:

The discharge/charge equation is an exponential function. Under normal conditions where the capacitor discharges or charges via a fixed resistance, the rate at which charge changes therefore also follows an exponential increase/decrease.

The clue when not to use is given by the question stating constant current. i.e. in order to supply a constant current, the load resistance cannot be fixed and must vary accordingly.

The time (-t) given by the fixed resistor discharge equation, is the time over which the discharge (or charge) occurs.