# Capacitors

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#1
Sorry I've been spamming the physics thread lately
A 1000 μF capacitor is charged with a battery and then discharged through a resistor. What resistance is required to cause the potential difference to drop to 5% in one minute?

If I use the equation I have time = 60 seconds but I am missing C, V, and so how do I find those out?
Last edited by Mlopez14; 4 weeks ago
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4 weeks ago
#2
C is given in the question.
You don't need to know the specific values of V and V0. Instead, consider what the equation is actually showing. The term e-t/RC gives the ratio between V and V0 after some time t, which you can see more clearly if you divide both sides of the equation by V0. What would the value of this ratio, V/V0, be, when the pd has dropped down to 5%?
Last edited by nzy; 4 weeks ago
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#3
(Original post by nzy)
C is given in the question.
You don't need to know the specific values of V and V0. Instead, consider what the equation is actually showing. The term e-t/RC gives the ratio between V and V0 after some time t, which you can see more clearly if you divide both sides of the equation by V0. What would the value of this ratio, V/V0, be, when the pd has dropped down to 5%?
I found C to be 0.001 F but I am not sure how to find that ratio since I am still missing resistance, which is what I need to found
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4 weeks ago
#4
(Original post by Mlopez14)
I found C to be 0.001 F but I am not sure how to find that ratio since I am still missing resistance, which is what I need to found
Take natural logs on both sides of the equation to put it into a form for algebraic manipulation. Then rearrange to give an expression for R before substituting values.
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#5
(Original post by uberteknik)
Take natural logs on both sides of the equation to put it into a form for algebraic manipulation. Then rearrange to give an expression for R before substituting values.
I haven't studied logarithms yet. But in that case I would still be missing both voltage and initial voltage
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4 weeks ago
#6
(Original post by Mlopez14)
I haven't studied logarithms yet. But in that case I would still be missing both voltage and initial voltage
You have obviously worked out how to write equations using latex though?

The question does not ask for the value of the initial and final voltages. It does give the final voltage as a ratio of the initial vs final if the expression is rearranged:

V = Voe-t/CR

V/Vo = e-t/CR

e.g. where V/Vo = 5% = 0.05 i.e. the absolute values are always in the same ratio to give 5%.

If you have not studied logarithms yet, then other solution methods will be well beyond the level of your current maths learning.

Definitely study logs before attempting to find the solution.

You will end up with an expression:

ln(V/Vo) = ln{e^-t/CR}

i.e. taking the log of e raised to the power of (-t/CR) leaves just the exponential (-t/CR) on the RHS:

ln(0.05) = -t/CR

rearranging for R as the subject:

R = -t/C*ln(0.05)

sub in the given component values and calculate ln(0.05) = -1.301

which gives:

R = -60/(0.001 * -1.301)

R = 46.12*103 ohms
1
#7
(Original post by uberteknik)
You have obviously worked out how to write equations using latex though?

The question does not ask for the value of the initial and final voltages. It does give the final voltage as a ratio of the initial vs final if the expression is rearranged:

V = Voe-t/CR

V/Vo = e-t/CR

e.g. where V/Vo = 5% = 0.05 i.e. the absolute values are always in the same ratio to give 5%.

If you have not studied logarithms yet, then other solution methods will be well beyond the level of your current maths learning.

Definitely study logs before attempting to find the solution.

You will end up with an expression:

ln(V/Vo) = ln{e^-t/CR}

i.e. taking the log of e raised to the power of (-t/CR) leaves just the exponential (-t/CR) on the RHS:

ln(0.05) = -t/CR

rearranging for R as the subject:

R = -t/C*ln(0.05)

sub in the given component values and calculate ln(0.05) = -1.301

which gives:

R = -60/(0.001 * -1.301)

R = 46.12*103 ohms
Thanks, I understood all of that.
Luckily, I am studying that quite soon.
On a side note, if I enter In(0.05), I get -2.99. Not sure why?
Nevertheless thank you
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4 weeks ago
#8
(Original post by Mlopez14)
Thanks, I understood all of that.
Luckily, I am studying that quite soon.
On a side note, if I enter In(0.05), I get -2.99. Not sure why?
Nevertheless thank you

Yes, we must take natural logs on both sides since the RHS has 'e' on the RHS, which then mkes the working easier. I erroneously took the base 10 logarithm 0f 0.05 on the LHS (-1.301) and mixed it up with the natural log of 'e' on the RHS which gave the wrong answer.

The correct answer should be 20K ohms rounded to 2 SF.

The answer would still be correct if I had kept the same log base on both sides. i.e. either used log10 on both sides or used loge on both sides and not mixed them together.

i.e. log10e(-t/CR) = 0.434(-t/CR)

whereas

loge(-t/CR) = -t/CR

Be warned!
Last edited by uberteknik; 4 weeks ago
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