B
Boki
Guest
Hi, All:
Could you please explain why Q=C*V?
Thanks.
Boki
Could you please explain why Q=C*V?
Thanks.
Boki
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Don Kelly
Guest
"Boki" <bokiteam@ms21.hinet.net> wrote in message
news:bfni3t$k7f@netnews.hinet.net...
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer
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Put it this way: C is defined as the ratio Q/VHi, All:
Could you please explain why Q=C*V?
Thanks.
Boki
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Don Kelly
dhky@peeshaw.ca
remove the urine to answer
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grahamk
Guest
"Boki" <bokiteam@ms21.hinet.net> wrote in message
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Beginners and Intermediate Electronics
http://homepage.ntlworld.com/g.knott/index.htm
news:bfni3t$k7f@netnews.hinet.net...
the charge( Q) is.Hi, All:
Could you please explain why Q=C*V?
Thanks.
Boki
The larger the Capacitor is, and the higher the Voltage is, then the larger
--
Beginners and Intermediate Electronics
http://homepage.ntlworld.com/g.knott/index.htm
J
Jonathan Kirwan
Guest
On Thu, 24 Jul 2003 17:50:00 +0100, "grahamk"
<g.knott@ntlworld.com> wrote:
capacitor, the higher the voltage the more charge you expect,
so:
q proportional to V
and for a given voltage, the bigger the capacitor is the more
charge you expect, so:
q proportional to c
There don't appear to be other obvious factors at this point,
so:
q proportional to c*v
or
q = k1 * c * v
with 'k1' being yet to be determined.
Or, rearranged, C = k2*Q/V.
Turns out, C is *defined* so that k2 is just 1.000 when Q is in
Coulombs and V is in Joules per Coulomb. Capacitance is in
Coulumb^2/Joule.
Add to this, an Ohm is a Joule-sec/Coulomb^2 (a Joule-sec is a
measure of angular momentum, for example, and is the units of
Planck's constant -- it's also just kg*(m/s)*m.)
Multiply units of R*C:
(Joule-sec/Coulomb^2) * (Coulomb^2/Joule)
and you get seconds! Good thing.
Since a Henry is just a Joule-sec^2/Coulomb^2 (a kind of
Ohm-second or Joule per amp^2), the SQRT(L*C) also works out to
seconds. Also a good thing. And L/R is also in seconds.
Another good thing.
Jon
<g.knott@ntlworld.com> wrote:
That's a dimensional analysis way of looking at it. For a given"Boki" <bokiteam@ms21.hinet.net> wrote in message
news:bfni3t$k7f@netnews.hinet.net...
Hi, All:
Could you please explain why Q=C*V?
Thanks.
Boki
The larger the Capacitor is, and the higher the Voltage is, then the larger
the charge( Q) is.
capacitor, the higher the voltage the more charge you expect,
so:
q proportional to V
and for a given voltage, the bigger the capacitor is the more
charge you expect, so:
q proportional to c
There don't appear to be other obvious factors at this point,
so:
q proportional to c*v
or
q = k1 * c * v
with 'k1' being yet to be determined.
Or, rearranged, C = k2*Q/V.
Turns out, C is *defined* so that k2 is just 1.000 when Q is in
Coulombs and V is in Joules per Coulomb. Capacitance is in
Coulumb^2/Joule.
Add to this, an Ohm is a Joule-sec/Coulomb^2 (a Joule-sec is a
measure of angular momentum, for example, and is the units of
Planck's constant -- it's also just kg*(m/s)*m.)
Multiply units of R*C:
(Joule-sec/Coulomb^2) * (Coulomb^2/Joule)
and you get seconds! Good thing.
Since a Henry is just a Joule-sec^2/Coulomb^2 (a kind of
Ohm-second or Joule per amp^2), the SQRT(L*C) also works out to
seconds. Also a good thing. And L/R is also in seconds.
Another good thing.
Jon