# Need L and C to be in series in order to have a RLC series circuit?

Swapnil
Do you only need L and C to be in series in order to have a RLC series circuit? (So where you put R doesn't matter right?).

doodle
Strictly speaking no, the R should be in series with L and C too. Then again, an R which is in parallel to the LC pair could be converted (with network transformation theorems) to being in series to it.

So I guess there is really no straight answer to this question.

antoker
Do you only need L and C to be in series in order to have a RLC series circuit? (So where you put R doesn't matter right?).

RLC series circuit implies that RL are in series & C is shunted to the ground, that again implies that we have a low-pass circuit of second order (capacitor provides an escape route for high frequency). Now interchanging R and L will have no affect on the circuit, since they are in series (circuit will have same fc etc.) But if you interchange L & C you will end up with a high-pass circuit, since the capacitor will exhibit high Xc for low frequencies and low Xc for high frequencies. So it does matter in how you arrange the L&C, but it doesn't matter in how you arrange the R&C.

P.S You can prove this by writing a transfer function for the both circuits (swapped R) using laplace and you'll end up the similar functions.

Mentor
Do you only need L and C to be in series in order to have a RLC series circuit? (So where you put R doesn't matter right?).

A series RLC circuit has all 3 elements in series. That's why it's called a "series RLC circuit" after all. Is there a context to this question?

Swapnil
A series RLC circuit has all 3 elements in series. That's why it's called a "series RLC circuit" after all. Is there a context to this question?
Its just that you have these equations for a series and a parallel RLC circuit so I was wondering if you have circuit which has any arbitrary configuration of a resistor, a capacitor, and an inductor, would you always be able to call it either a series or parallel RLC circuit?