I have two very simple questions about capacitors. My answers differ from
those on the answer sheet. Can someone please explain where I have made a
mistake.

1. Two capacitors C1=10 microfarad and C2=15 microfarad are connected in
series and a potential difference of 500V put across them. Determine

a. the charge stored on each capacitor
b. the potential difference across each capacitor

When I simulate this circuit in Multisim, a voltmeter across each capacitor
shows the same potential difference of 250V across each capacitor. From the
formula for charge on a capacitor q = CV, I get

q1 = 10 x 10^-6 x 250 = 2.50 x 10^-3 Coulomb charge on C1
q2 = 15 x 10^-6 x 250 = 3.75 x 10^-3 Coulomb charge on C2

The book answers are

3 x 10^-3 Coulomb charge stored on each capacitor.
Potential difference 300V on C1
Potential difference 200V on C2


2. Three 10 microfarad capacitors are connected in series. The combination
is connected to a 1.5V battery. Calculate the charge stored in each
capacitor and the total charge stored.

The three capacitors can be replaced by one equivalent capacitor of 10/3
microfarad.

The charge on this single equivalent capacitor is q = CV so
q = 10/3 x 10^-6 x 1.5 = 5 x 10 ^-6 Coulomb total charge, which is correct.

The voltage drop across each capacitor is 0.5V (confirmed by Multisim)

The charge on each capacitor is given by q = CV so the charge on the first
capacitor is

10 x 10^-6 x 0.5 = 5 x 10^-6 Coulomb on the first and hence each capacitor.

The correct answer is 1.7 x 10^-6 Coulomb.

My answers are nonsense, but in terms of the equations they "look" OK

Any help much appreciated.

Brad

"Brad Cooper" <Brad.Cooper_17@bigpond.com> wrote in message
news:lzzzi.25563$4A1.13715@news-server.bigpond.net.au...
>I have two very simple questions about capacitors. My answers differ from
>those on the answer sheet. Can someone please explain where I have made a
>mistake.
>
> 1. Two capacitors C1=10 microfarad and C2=15 microfarad are connected in
> series and a potential difference of 500V put across them. Determine
>
> a. the charge stored on each capacitor
> b. the potential difference across each capacitor
>
> When I simulate this circuit in Multisim, a voltmeter across each
> capacitor shows the same potential difference of 250V across each
> capacitor. From the formula for charge on a capacitor q = CV, I get
>
> q1 = 10 x 10^-6 x 250 = 2.50 x 10^-3 Coulomb charge on C1
> q2 = 15 x 10^-6 x 250 = 3.75 x 10^-3 Coulomb charge on C2
>
> The book answers are
>
> 3 x 10^-3 Coulomb charge stored on each capacitor.
> Potential difference 300V on C1
> Potential difference 200V on C2
>
Remember - for the series combination, the charge is the same on each
capacitor (total charge = charge on C1 = charge on C2, etc). Total
capacitance in your circuit is C1C2/(C1+C2) = 6 uF. The total charge, and
the charge on each capacitor, is 6e-6*500=3e-3 Coulomb.
Rearranging the terms in q=CV to find the voltage across each capacitor:
V=q/C, which results in the Vc1=300V and Vc2=200V.

> 2. Three 10 microfarad capacitors are connected in series. The combination
> is connected to a 1.5V battery. Calculate the charge stored in each
> capacitor and the total charge stored.
>
> The three capacitors can be replaced by one equivalent capacitor of 10/3
> microfarad.
>
> The charge on this single equivalent capacitor is q = CV so
> q = 10/3 x 10^-6 x 1.5 = 5 x 10 ^-6 Coulomb total charge, which is
> correct.
>
> The voltage drop across each capacitor is 0.5V (confirmed by Multisim)
>
> The charge on each capacitor is given by q = CV so the charge on the first
> capacitor is
>
> 10 x 10^-6 x 0.5 = 5 x 10^-6 Coulomb on the first and hence each
> capacitor.
>
> The correct answer is 1.7 x 10^-6 Coulomb.

No, it isn't. Everything was fine up to this point. It looks like whoever
arrived at that answer for the book (probably some over-worked grad student)
jumped from the given series combination to a parallel combination of three
10uF capacitors with voltage of 0.1667 volts across them. Actually, what
they probably did was divide the charge (5e-6 Coulomb) by 3, but a bad leap,
none the less.
The correct answer is, as you found, 5e-6 Coulomb on each capacitor.

The moral of the story is that you can't always trust the answers in the
book.

> My answers are nonsense, but in terms of the equations they "look" OK
>
> Any help much appreciated.
>
> Brad
I hope this helps.

Richard

Richard Seriani wrote:

SNIP
>>
>> The correct answer is 1.7 x 10^-6 Coulomb.
>
> No, it isn't. Everything was fine up to this point. It looks like whoever
> arrived at that answer for the book (probably some over-worked grad student)
> jumped from the given series combination to a parallel combination of three
> 10uF capacitors with voltage of 0.1667 volts across them. Actually, what
> they probably did was divide the charge (5e-6 Coulomb) by 3, but a bad leap,
> none the less.

I agree with Richard, although that grad
student must have understood that the
charges on series capacitors are equal
in order to arrive at the correct answer
to #1.

Chuck

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On Fri, 24 Aug 2007 11:47:29 GMT, "Brad Cooper"
<Brad.Cooper_17@bigpond.com> wrote:

>I have two very simple questions about capacitors. My answers differ from
>those on the answer sheet. Can someone please explain where I have made a
>mistake.
>
>1. Two capacitors C1=10 microfarad and C2=15 microfarad are connected in
>series and a potential difference of 500V put across them. Determine
>
>a. the charge stored on each capacitor
>b. the potential difference across each capacitor
>
>When I simulate this circuit in Multisim, a voltmeter across each capacitor
>shows the same potential difference of 250V across each capacitor. From the
>formula for charge on a capacitor q = CV, I get
>
>q1 = 10 x 10^-6 x 250 = 2.50 x 10^-3 Coulomb charge on C1
>q2 = 15 x 10^-6 x 250 = 3.75 x 10^-3 Coulomb charge on C2
>
>The book answers are
>
>3 x 10^-3 Coulomb charge stored on each capacitor.
>Potential difference 300V on C1
>Potential difference 200V on C2

I'm surprised that Multisim will even run this situation; most Spice
programs won't, because the answer is actually mathematically
indeterminate. I'm not a bit surprised that Multisim gave an idiotic
answer... it tends to do that.

The book is right *assuming* zero initial charges and no leakages.






>
>
>2. Three 10 microfarad capacitors are connected in series. The combination
>is connected to a 1.5V battery. Calculate the charge stored in each
>capacitor and the total charge stored.
>
>The three capacitors can be replaced by one equivalent capacitor of 10/3
>microfarad.
>
>The charge on this single equivalent capacitor is q = CV so
>q = 10/3 x 10^-6 x 1.5 = 5 x 10 ^-6 Coulomb total charge, which is correct.
>
>The voltage drop across each capacitor is 0.5V (confirmed by Multisim)
>
>The charge on each capacitor is given by q = CV so the charge on the first
>capacitor is
>
>10 x 10^-6 x 0.5 = 5 x 10^-6 Coulomb on the first and hence each capacitor.

That looks right, again assuming zero initial charges.


>
>The correct answer is 1.7 x 10^-6 Coulomb.

That doesn't.

John

John Larkin wrote:

>
> I'm surprised that Multisim will even run this situation; most Spice
> programs won't, because the answer is actually mathematically
> indeterminate.

Why mathematically indeterminate, John?

The charge on any capacitor in series is
the same as the total charge on the
combination (calculation of total charge
trivial).

The voltage across C1 is the voltage
across both capacitors times (total
capacitance/C1).

So the book is wrong on #1 as well as
#2, as explained by Richard.

Chuck

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Chuck wrote:

>
> So the book is wrong on #1 as well as #2, as explained by Richard.
>

I should have said partly correct and
partly in error on each part.

Hate it when that happens.

Chuck

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On Fri, 24 Aug 2007 11:47:29 +0000, Brad Cooper wrote:

> I have two very simple questions about capacitors. My answers differ from
> those on the answer sheet. Can someone please explain where I have made a
> mistake.
>
> 1. Two capacitors C1=10 microfarad and C2=15 microfarad are connected in
> series and a potential difference of 500V put across them. Determine
>
> a. the charge stored on each capacitor
> b. the potential difference across each capacitor
>
> When I simulate this circuit in Multisim, a voltmeter across each capacitor
> shows the same potential difference of 250V across each capacitor. From the
> formula for charge on a capacitor q = CV, I get
>
> q1 = 10 x 10^-6 x 250 = 2.50 x 10^-3 Coulomb charge on C1
> q2 = 15 x 10^-6 x 250 = 3.75 x 10^-3 Coulomb charge on C2
>
> The book answers are
>
> 3 x 10^-3 Coulomb charge stored on each capacitor.
> Potential difference 300V on C1
> Potential difference 200V on C2

The book is correct.

For two capacitors in series, the charge on each has to be identical. The
amount of charge fed into one end equals the charge which moves from C1 to
C2 which equals the charge coming out of the other end.

Did you actually use 10uF and 15uF capacitors in the simulation? Or did
you e.g. use two identical capacitors labelled differently? [I haven't
used Multisim, but some schematic programs make it all too easy to add a
label which doesn't correspond to the value.]

> 2. Three 10 microfarad capacitors are connected in series. The combination
> is connected to a 1.5V battery. Calculate the charge stored in each
> capacitor and the total charge stored.
>
> The three capacitors can be replaced by one equivalent capacitor of 10/3
> microfarad.
>
> The charge on this single equivalent capacitor is q = CV so
> q = 10/3 x 10^-6 x 1.5 = 5 x 10 ^-6 Coulomb total charge, which is correct.
>
> The voltage drop across each capacitor is 0.5V (confirmed by Multisim)
>
> The charge on each capacitor is given by q = CV so the charge on the first
> capacitor is
>
> 10 x 10^-6 x 0.5 = 5 x 10^-6 Coulomb on the first and hence each capacitor.

.... and hence in total.

> The correct answer is 1.7 x 10^-6 Coulomb.

No, 5uC.

10uF * 0.5V = 5uC
10/3uF * 1.5V = 5uC

> My answers are nonsense, but in terms of the equations they "look" OK
>
> Any help much appreciated.

When capacitors are connected in series, the amount of charge stored by
each one is identical, and is also equal to the charge stored by the
entire chain.

Brad Cooper wrote:

> I have two very simple questions about capacitors. My answers differ from
> those on the answer sheet. Can someone please explain where I have made a
> mistake.
>
> 1. Two capacitors C1=10 microfarad and C2=15 microfarad are connected in
> series and a potential difference of 500V put across them. Determine
>
> a. the charge stored on each capacitor
> b. the potential difference across each capacitor
>
> When I simulate this circuit in Multisim, a voltmeter across each
> capacitor
> shows the same potential difference of 250V across each capacitor. From
> the formula for charge on a capacitor q = CV, I get
>
> q1 = 10 x 10^-6 x 250 = 2.50 x 10^-3 Coulomb charge on C1
> q2 = 15 x 10^-6 x 250 = 3.75 x 10^-3 Coulomb charge on C2
>
> The book answers are
>
> 3 x 10^-3 Coulomb charge stored on each capacitor.
> Potential difference 300V on C1
> Potential difference 200V on C2
>
>
> 2. Three 10 microfarad capacitors are connected in series. The combination
> is connected to a 1.5V battery. Calculate the charge stored in each
> capacitor and the total charge stored.
>
> The three capacitors can be replaced by one equivalent capacitor of 10/3
> microfarad.
>
> The charge on this single equivalent capacitor is q = CV so
> q = 10/3 x 10^-6 x 1.5 = 5 x 10 ^-6 Coulomb total charge, which is
> correct.
>
> The voltage drop across each capacitor is 0.5V (confirmed by Multisim)
>
> The charge on each capacitor is given by q = CV so the charge on the first
> capacitor is
>
> 10 x 10^-6 x 0.5 = 5 x 10^-6 Coulomb on the first and hence each
> capacitor.
>
> The correct answer is 1.7 x 10^-6 Coulomb.
>
> My answers are nonsense, but in terms of the equations they "look" OK

The question is nonsense. If the capacitors are ideal, the answers depend on
the initial charge, which is not given. If the capacitors are not ideal,
there is a non-zero leakage current and the circuit reduces to a voltage
divider with unknown resistances.

>
> Any help much appreciated.
>
> Brad

HTH,

--
Niels Diepeveen

"Niels Diepeveen"

> The question is nonsense. If the capacitors are ideal, the answers depend
> on
> the initial charge, which is not given.

** So there isn't any.

Fool.



........ Phil

On Fri, 24 Aug 2007 20:21:22 +0100, Nobody <nobody@nowhere.com> wrote:

>On Fri, 24 Aug 2007 11:47:29 +0000, Brad Cooper wrote:
>
>> I have two very simple questions about capacitors. My answers differ from
>> those on the answer sheet. Can someone please explain where I have made a
>> mistake.
>>
>> 1. Two capacitors C1=10 microfarad and C2=15 microfarad are connected in
>> series and a potential difference of 500V put across them. Determine
>>
>> a. the charge stored on each capacitor
>> b. the potential difference across each capacitor
>>
>> When I simulate this circuit in Multisim, a voltmeter across each capacitor
>> shows the same potential difference of 250V across each capacitor. From the
>> formula for charge on a capacitor q = CV, I get
>>
>> q1 = 10 x 10^-6 x 250 = 2.50 x 10^-3 Coulomb charge on C1
>> q2 = 15 x 10^-6 x 250 = 3.75 x 10^-3 Coulomb charge on C2
>>
>> The book answers are
>>
>> 3 x 10^-3 Coulomb charge stored on each capacitor.
>> Potential difference 300V on C1
>> Potential difference 200V on C2
>
>The book is correct.
>
>For two capacitors in series, the charge on each has to be identical. The
>amount of charge fed into one end equals the charge which moves from C1 to
>C2 which equals the charge coming out of the other end.
>
>Did you actually use 10uF and 15uF capacitors in the simulation? Or did
>you e.g. use two identical capacitors labelled differently? [I haven't
>used Multisim, but some schematic programs make it all too easy to add a
>label which doesn't correspond to the value.]
>
>> 2. Three 10 microfarad capacitors are connected in series. The combination
>> is connected to a 1.5V battery. Calculate the charge stored in each
>> capacitor and the total charge stored.
>>
>> The three capacitors can be replaced by one equivalent capacitor of 10/3
>> microfarad.
>>
>> The charge on this single equivalent capacitor is q = CV so
>> q = 10/3 x 10^-6 x 1.5 = 5 x 10 ^-6 Coulomb total charge, which is correct.
>>
>> The voltage drop across each capacitor is 0.5V (confirmed by Multisim)
>>
>> The charge on each capacitor is given by q = CV so the charge on the first
>> capacitor is
>>
>> 10 x 10^-6 x 0.5 = 5 x 10^-6 Coulomb on the first and hence each capacitor.
>
>... and hence in total.
>
>> The correct answer is 1.7 x 10^-6 Coulomb.
>
>No, 5uC.
>
>10uF * 0.5V = 5uC
>10/3uF * 1.5V = 5uC
>
>> My answers are nonsense, but in terms of the equations they "look" OK
>>
>> Any help much appreciated.
>
>When capacitors are connected in series, the amount of charge stored by
>each one is identical, and is also equal to the charge stored by the
>entire chain.

Yikes, I'm so used to reflexively thinking that "charge is conserved",
it seems counter-intuitive that each capacitor has the same charge as
*all of the capacitors in series*, but it does. After all, each cap
saw some amount of ampere-seconds to charge it up, and the entire
string, as a black box, saw the same charge.

So do the experiment: three 10u caps, discharged. Connect them in
series to a 1.5 volt battery. The string is 3.3uF, times 1.5 volts,
for 5 uC net. Each cap is 10 uF charged to 0.5 volts, so each cap also
stores 5 uC.

Now disconnect them and reconnect in parallel. Now the available
charge is 15 uC. Cool.

John

On Sat, 25 Aug 2007 15:52:51 -0700, John Larkin wrote:

>>When capacitors are connected in series, the amount of charge stored by
>>each one is identical, and is also equal to the charge stored by the
>>entire chain.
>
> Yikes, I'm so used to reflexively thinking that "charge is conserved",
> it seems counter-intuitive that each capacitor has the same charge as
> *all of the capacitors in series*, but it does. After all, each cap
> saw some amount of ampere-seconds to charge it up, and the entire
> string, as a black box, saw the same charge.

I know what you mean.

It's all too easy to think of charging a capacitor as filling a tank
with little blobs of charge, so it seems natural to add the charges on the
two capacitors.

But what's really at issue is the amount of charge on the "wrong" side of
the dielectric, i.e. an excess of electrons on one side and a deficit on
the other. Charged or discharged, the total number of electrons in the
capacitor will be constant (and equal to the number of protons).

On Sat, 25 Aug 2007 15:52:51 -0700, John Larkin wrote:
> On Fri, 24 Aug 2007 20:21:22 +0100, Nobody <nobody@nowhere.com> wrote:
>
>>When capacitors are connected in series, the amount of charge stored by
>>each one is identical, and is also equal to the charge stored by the
>>entire chain.
>
> Yikes, I'm so used to reflexively thinking that "charge is conserved",
> it seems counter-intuitive that each capacitor has the same charge as
> *all of the capacitors in series*, but it does. After all, each cap saw
> some amount of ampere-seconds to charge it up, and the entire string, as
> a black box, saw the same charge.
>
> So do the experiment: three 10u caps, discharged. Connect them in series
> to a 1.5 volt battery. The string is 3.3uF, times 1.5 volts, for 5 uC
> net. Each cap is 10 uF charged to 0.5 volts, so each cap also stores 5
> uC.
>
> Now disconnect them and reconnect in parallel. Now the available charge
> is 15 uC. Cool.

Sure - it's a Cockroft-Walton in reverse.

Maybe it's because the charge is "stored" in the dielectric?

Thanks,
Rich