# Does a LC filter filters the voltage or the current or both?

The above image shows a simple low pass LC filter with an AC input source. Imagine that a resistor is being connected at the output as the load.

Does it filter the

1) high frequency voltages

2) high frequency currents

3) or both of them?

• Your question does not make sense. Is a two resistor-network a voltage divider or is it a current divider? This depends on what is connected to it. May 10 '17 at 13:56
• "Is a two resistor-network a voltage divider or is it a current divider?" When the voltage is divided between the two resistors, so does the current. May 10 '17 at 14:01
• It's a 2nd order LC filter with undefined R - voltages and currents ALWAYS get filtered to different levels depending on the frequency of the applied input energies. Depending on the frequency, there may be very little noticeable difference between input and output, or there maybe a marked reduction or there maybe a massive amplification. Take your pick. May 10 '17 at 14:02
• @Taven: That's what I wanted to say. We talk about voltage dividers knowing this is a simplification. For the original question, to know if we can live with that simplification or have to consider the load, too, depends on this load. May 10 '17 at 14:06
• @Janka it's an inductor not a resistor. May 10 '17 at 14:21

Does a LC filter filters the voltage or the current or both?

It all depends on what you mean by "filter".

At low frequencies, the output voltage follows the input voltage because we can assume the inductor is a short circuit and the capacitor is an open circuit.

At the resonant frequency, L and C form a series tuned circuit across the input and act like a short circuit so, around resonance, the input is heavily loaded and large currents are taken. These large currents produce output voltage magnification.

At high frequencies the inductor starts to act as an open circuit and the capacitor starts to act as a short circuit and input voltages are heavily attenuated.

That's the simple version of events.

Here's a typical spectrum of a 2nd order (RLC) low pass filter: -

Depending on Q the output can be very peaky at resonance. Higher Q means higher peaks. Q is determined, in part by the load resistor. If R is high value then Q will be high; if R is low value then Q (and peaking will be low).

Here's a link to another stack exchange question and answer that goes a bit deeper.

simulate this circuit – Schematic created using CircuitLab

With a purely resistive, ideal load, the current through the load follows Ohm's Law.

V = IR

So if the magnitude of V is reduced at higher frequencies, and R is constant, I must necessarily be reduced as well, according to:

I = V/R

In a purely resistive load, there is no phase offset between voltage and current - the current follows the voltage.

This assumes an ideal circuit and is greatly oversimplified from real-world behavior.