# A basic math question about the current in a capacitor

1. Dec 4, 2013

### Adir_Sh

The current in a capacitor is calculated from

$i_{C}(t)=C \dfrac{dv_{C}(t)}{dt}$​

when $i_{C}(t)$ is the current in the capacitor, $t$ is the time variable, $C$ is the capacitance and $v_{C}(t)$ is the capacitor's voltage.
This equation can also be expressed, using $\omega t$ as the variable, as

$i_{C}(\omega t)=\omega C \dfrac{dv_{C}(\omega t)}{d(\omega t)}$​

Now my basic question is why there's a multiplication in $\omega$? That is, why is it correct to multiply when changing in independent variable and multiplying in by frequency $\omega$? Seemingly, the new variable, $\omega t$, is derived in the same manner as $t$ is derived when the function uses only $t$ as the independent variable.
I don't quite understand this little issue. Hope someone could explain that in a sentence.

Thanks in advance! :tongue:

2. Dec 4, 2013

### tiny-tim

Hi Adir_Sh!
chain rule

dvC/dt

= dvC/d(ωt)*d(ωt)/dt

= dvC/d(ωt)*ω

3. Dec 5, 2013

### rude man

Realize that d(wt) = w dt and you have your answer.

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