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'Calculus question'
2012\01\29@170252
by
jim
I have a question. Can anyone (or more than one), give me their answer to
this question...
If there is a current of 10ma charging a 1uF capacitor, what is the voltage
across the capacitor at 100ms?
I have calculated my answer twice, and come up with the same answer twice.
But it just doesn't seem right to me.
So, I thought I'd get an independent, impartial answer to verify or deny my
answer.
If you choose, put your solution to the answer down too, so I can follow
your thinking. I am not in school, and this is not a homework question. I am just curious. It's actually a
problem in an electronics book, and I wanted to answer it, but I don't think my answer is correct, and the answer isn't
in the book.
Thanks and Regards,
Jim
2012\01\29@171528
by
Roy
2012\01\29@171742
by
Wouter van Ooijen
On 29/1/2012 11:02 PM, jim wrote:
> I have a question. Can anyone (or more than one), give me their answer to
> this question...
>
> If there is a current of 10ma charging a 1uF capacitor, what is the voltage
> across the capacitor at 100ms?
>
> I have calculated my answer twice, and come up with the same answer twice..
> But it just doesn't seem right to me.
> So, I thought I'd get an independent, impartial answer to verify or deny my
> answer.
>
> If you choose, put your solution to the answer down too, so I can follow
> your thinking. I am not in school, and
> this is not a homework question. I am just curious. It's actually a
> problem in an electronics book, and I wanted
> to answer it, but I don't think my answer is correct, and the answer isn't
> in the book.
1A into (or out of) a 1F capacitor for 1 second gives a deltaV of 1V.
Your capacitor is only e6 F, your current is e2, your time e1.
I am a (part time) teacher, so I leave the rest as en exercise for the student.
If your answer seems to high for you, realise that your capacitor is small. 1mA for 1ms into a 0F capacitor will give an interesting effect (don't try this at home).

Wouter van Ooijen
 
Van Ooijen Technische Informatica: http://www.voti.nl
consultancy, development, PICmicro products
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2012\01\29@172016
by
jim
It's a theoretical question only, so the starting voltage is zero, and the
resistances are zero.
{Original Message removed}
2012\01\29@172638
by
William \Chops\ Westfield
On Jan 29, 2012, at 2:02 PM, jim wrote:
> If there is a current of 10ma charging a 1uF capacitor, what is the voltage
> across the capacitor at 100ms?
v(T) = 1/C (integrate(i(t), dt, t0, T)
= 1e6 * 0.01 * .1
= 1e3 V
Is that what you got? Sanity check: if we has a 1e3 V power supply, then to limit current to 10mA we would use a 100k ohm resistor, and the RC time constant would be
1e6*1e5 = 0.1s. Yep. (This would not result in a constant current of 10mA, and the capacitor voltage would not be 1000V after 1 time constant, but that it is the correct order of magnitude is reassuring.)
It does seem surprising, since 1uF and 10mA are values you'd expect to find in a real circuit, and 1e5 V isn't. However, pumping constant current into a cap is not trivial to actually accomplish. Most real circuits are closer to voltage sources with resistance, and current into caps quickly falls to small amounts.
BillW
2012\01\29@173517
by
peter green

jim wrote:
> I have a question. Can anyone (or more than one), give me their answer to
> this question...
>
> If there is a current of 10ma charging a 1uF capacitor, what is the voltage
> across the capacitor at 100ms?
> You don't really need calculus for this one (calculus would only been needed if charging
with a voltage dependent current (e,g. charging from a voltage source through a resistor).
Assuming the charge on the capacitor is zero at time zero (you didn't specify any boundry
conditions)
Using SI units through (I could put unit symbols on but given that this is a plain text email
they would probablly confuse more than help)
Q = (10x10^3)*(100*10^3)=10^3
C = 10^6
C = Q/V
(10^6)=(10^3)/V
V=1000
Of course in practice you will almost never be charging from a constant current source...
{Quote hidden}> I have calculated my answer twice, and come up with the same answer twice..
> But it just doesn't seem right to me.
> So, I thought I'd get an independent, impartial answer to verify or deny my
> answer.
>
> If you choose, put your solution to the answer down too, so I can follow
> your thinking. I am not in school, and
> this is not a homework question. I am just curious. It's actually a
> problem in an electronics book, and I wanted
> to answer it, but I don't think my answer is correct, and the answer isn't
> in the book.
>
> Thanks and Regards,
>
> Jim
>
>
2012\01\29@174014
by
jim
Bill,
Yes, that is exactly what I got. 1kV. However, when I put this circuit into
SPICE (actually TINATI), the answer that SPICE comes up with is 9.09kV.
I'm not sure why this is. Is it a mistake in my (our) calculations, or is
it maybe with Zero ohms of series resistance, SPICE is confused?
Thanks and Regards,
Jim
{Original Message removed}
2012\01\29@180736
by
Spehro Pefhany
At 05:40 PM 1/29/2012, you wrote:
>Bill,
>
>Yes, that is exactly what I got. 1kV. However, when I put this circuit into
>SPICE (actually TINATI), the answer that SPICE comes up with is 9.09kV.
>I'm not sure why this is. Is it a mistake in my (our) calculations, or is
>it maybe with Zero ohms of series resistance, SPICE is confused?
>
>Thanks and Regards,
>
>Jim
I'll assume you've grounded one side of the circuit.
The problem is that SPICE will not be able to calculate the initial bias
point  it has no stable bias point. Try BOTH telling it to skip that calculation
and put a very large value resistor in parallel with the cap (maybe 1000G ohms)
Best regards,
2012\01\29@180949
by
YES NOPE9
Remember that in the real world a constant current source may fail at 5V or 100V ( or whatever ).
Before failing, the constant current source may stop being constant. The capacitor may stop being a capacitor.
Maybe SPICE does not believe there a voltages higher than .999KV and goes crazy.
Try a series resistance of 1 ohm. This will introduce an error factor of maybe .01V .
99guspuppet
{Quote hidden}> On Jan 29, 2012, at 3:40 PM, jim wrote:
>
>
> Bill,
>
> Yes, that is exactly what I got. 1kV. However, when I put this circuit into
> SPICE (actually TINATI), the answer that SPICE comes up with is 9.09kV.
> I'm not sure why this is. Is it a mistake in my (our) calculations, or is
> it maybe with Zero ohms of series resistance, SPICE is confused?
>
> Thanks and Regards,
>
> Jim
>
> {Original Message removed}
2012\01\29@185621
by
Oli Glaser
On 29/01/2012 23:07, Spehro Pefhany wrote:
> I'll assume you've grounded one side of the circuit.
>
> The problem is that SPICE will not be able to calculate the initial bias
> point  it has no stable bias point. Try BOTH telling it to skip that
> calculation
> and put a very large value resistor in parallel with the cap (maybe 1000G ohms)
I agree, that should work.
You could also set an initial condition for the capacitor node e.g:
..ic V(n001) = 0
'Calculus question'
2012\02\01@171024
by
Barry Gershenfeld
> Remember that in the real world a constant current source may fail at 5V
or 100V ( or whatever ).
This happens when the capacitor voltage gets near the constant current's
supply voltage. So model the current source at something much larger
(100kV) and a series resistor that limits the current to 10mA
2012\02\02@174059
by
Electron
At 23.02 2012.01.29, you wrote:
>I have a question. Can anyone (or more than one), give me their answer to
>this question...
>
>If there is a current of 10ma charging a 1uF capacitor, what is the voltage
>across the capacitor at 100ms?
Volt = seconds * Ampere / Farad
so
1000V = 0.1s * 0.01A / 0.000001F
but in practice it's not easy to sustain 10mA "indefinitely".
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