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I have two very related questions:

In the derivation of power in a circuit we can do the following:

\$P=\frac{dw}{dt}=\frac{d(QV)}{dt}=Q\frac{dV}{dt}+V\frac{dQ}{dt}\$

and then assuming that \$V\$ is constant:

\$P=V\frac{dQ}{dt}=VI\$

This formula holds in every case (that I can think of) in electronics, but why can we assume V is constant, even if we have an ac signal? Furthermore if we are assuming that V is constant why do we not assume that Q is constant also?

Secondly: When deriving the equation for energy stored in a capacitor you can work out the work done to move charge from one side plate to the other. But in the act of removing charge from one plate, you will change the potential between the plates, so why can we assume that the potential is constant when moving this charge from one plate to another. (the charge is usually infinitesimal (\$dQ\$) and the energy \$dU=VdQ\$ is then integrated over)

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  • \$\begingroup\$ "...why do we not assume that Q is constant also?" We don't make any assumptions about Q. If it's constant, then I = 0 and everything still works. Interesting question, though. \$\endgroup\$
    – Greg d'Eon
    Commented Feb 9, 2015 at 15:21
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    \$\begingroup\$ For AC signals you use the rms voltage or current. Root-mean-square integrates and averages to convert an AC signal to an equivalent DC signal with respect to power consumption. Then you can assume it is constant, however many times 'we' don't assume voltage or current is constant. \$\endgroup\$
    – HKOB
    Commented Feb 9, 2015 at 17:17
  • \$\begingroup\$ @Kynit I know that we don't make any assumptions about Q, that is part of my question, why are we making assumptions about V but not Q, it seems that we should be making assumptions of neither or both. \$\endgroup\$ Commented Feb 10, 2015 at 12:33
  • \$\begingroup\$ Oh, I just realized what's going on here. I'll write up an answer. \$\endgroup\$
    – Greg d'Eon
    Commented Feb 10, 2015 at 12:53

1 Answer 1

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One of your equations is incorrect: the energy in a capacitor is \$ \frac{1}{2}QV\$. Then, the power is

\$P = \frac{dw}{dt} = \frac{d}{dt}(\frac{1}{2}QV) = \frac{1}{2}[Q\frac{dV}{dt} + V\frac{dQ}{dt}]\$

Then, since V = \$\frac{Q}{C}\$, \$\frac{dV}{dt} = \frac{1}{C}\frac{dQ}{dt}\$, so

\$ P = \frac{1}{2}[\frac{Q}{C}\frac{dQ}{dt} + V\frac{dQ}{dt}] = \frac{1}{2}[V\frac{dQ}{dt} + V\frac{dQ}{dt}] = IV \$

with no assumptions about constant voltage.

I haven't thought about your second question, but I think the same idea will help out there too.

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  • \$\begingroup\$ No, it's \$\frac{1}{2}CV^2\$... \$\endgroup\$
    – Greg d'Eon
    Commented Feb 10, 2015 at 13:49
  • \$\begingroup\$ ...and power is change in energy over time. I followed OP's attempt above. \$\endgroup\$
    – Greg d'Eon
    Commented Feb 10, 2015 at 13:49
  • \$\begingroup\$ Nevermind, the Q confused me. \$\endgroup\$
    – Matt Young
    Commented Feb 10, 2015 at 13:50
  • \$\begingroup\$ @Kynit I can see this holds for a capacitor but what about in the general case? When work done (by a battery say) is QV \$\endgroup\$ Commented Feb 10, 2015 at 13:52
  • \$\begingroup\$ Hmm. I think it's easier for me to think of the problem in reverse: since you know that \$P = IV\$, if you know how V depends on Q, you can find the amount of energy from your two equations. Your question is then just applying that in reverse. I'm not totally sure about all of the details, though. \$\endgroup\$
    – Greg d'Eon
    Commented Feb 10, 2015 at 13:56

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