# Whats the relationship between capacitors' ability to filter high frequencies and i=C dv/dt

Whats the relationship between capacitors ability to filter out high frequencies and the equation i=C dv/dt?

Thanks

• Something to think about: capacitors can also be used to filter low frequencies. Also, I would suggest reading the Wikipedia page and meditating a bit, then coming here if your question isn't answered. Jun 2, 2017 at 3:34
• Think about what the equation is telling you, dv/dt means a rate of change of voltage with respect to time, lower rate of change (for example lower frequency) means lower current and viceversa, the extreme case is DC where dv/dt=0 so the current must be zero
– S.s.
Jun 2, 2017 at 3:37
• First draw the schematic of a single pole lowpass RC filter. Then manually 'play filter' by stepping through what happens when you apply a signal to it. It can be very instructive to program a spreadsheet to do this, far more instructive at your stage of education than simply putting the circuit into SPICE and seeing what happens, though this latter is also useful. Apply different frequencies. Repeat until realisation dawns. The key here is for you to do the maths. You need i=Cdv/dt to make the maths work, and you see what happens in the simulation. Jun 2, 2017 at 5:29
• If the amplitude of the sinusoidal voltage across the capacitor is 1V and the frequency is $\small \omega$, the current will be $i=C\frac{d}{dt}sin(\omega t)=\omega C\:cos(\omega t)$. So the current amplitude ($\small=\omega C$) will be zero when $\small \omega =0$, and will increase as $\omega$ increases.
– Chu
Jun 2, 2017 at 6:59