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

Thanks

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    \$\begingroup\$ 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. \$\endgroup\$ – uint128_t Jun 2 '17 at 3:34
  • \$\begingroup\$ 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 \$\endgroup\$ – S.s. Jun 2 '17 at 3:37
  • \$\begingroup\$ 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. \$\endgroup\$ – Neil_UK Jun 2 '17 at 5:29
  • \$\begingroup\$ 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. \$\endgroup\$ – Chu Jun 2 '17 at 6:59
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A filter cannot rely on a capacitor alone. It has to work in conjunction with another component, usually a resistor for example. The impedance of a capacitor is related to the formula I = C dv/dt in that if dv/dt becomes larger (a larger change in voltage with respect to time) then the current also becomes larger.

Given that a sine wave of a higher frequency has a larger dv/dt compared to a a sine wave of a lower frequency, you can probably recognize that the current increases for a larger frequency.

If a resistor and capacitor are used to form a potential divider like this: -

enter image description here

You might be able to recognize that for higher frequencies, more current flows into the capacitor than for lower frequencies. This means that the resistor "drops" more signal voltage at higher frequencies due to that increased current flow and, this means that the output signal amplitude is smaller when higher frequencies are presented to the input.

This circuit is a low-pass filter but if you swap the positions of R and C you get a high-pass filter.

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