LR/RC equivalent help

[SeanHatch]

If I have an Inductor, L, and a resistor, R, in series, how can I accomplish this same impedance with a resistor and a capacitor? I have no ideas, I've been working at this for an hour. Hint?

 

[phalanx]

In your textbook you will find a formula for the impedance of a simple RL circuit as well as a simple RC circuit. Set them equal to each other and solve using an arbitrary but realistic value for C.

 

[mstechca]

If I have an Inductor, L, and a resistor, R, in series, how can I accomplish this same impedance with a resistor and a capacitor?

Here is an answer that could help you:

You will need to figure out the frequency of the RC network, then find the reactance of the capacitor. That can be used as resistance when the circuit is running.

Do the same with the RL network, but find the reactance of the inductor.

Now the math is simple:

add the value of resistor and the reactance of the inductor to get the answer to the RL network.

add the value of resistor and the reactance of the capacitor to get the answer to the RC network.

Please understand that the reactance values are valid when the circuit using the RC and RL networks are in action.

If you don't want to make the circuit with these networks (or you wanted to add the circuit directly to a lamp), then there is no answer to your question, because a capacitor doesn't function the same as a resistor or an inductor.

Also, it may be impossible to measure resistance of an inductor because the value can be so small, most available DMM's are not designed for extra small numbers ( < 1 ohm).

 

[Roff]

Mstechca, you need to revert to your previous signature. It was very appropriate. Believe it or not, the impedance of an RL series network is not equal to the sum of the individual impedances.
Sean, are the resistor and capacitor also in series?

 

[SeanHatch]

It doesn't matter. I'm working out a transfer function, and while the RL series combo is much more intuitive, I've got to use capacitors.

:roll:


So, can I obtain the frequency(w) from the RL series combo, and then use that to find an equivalent impedance with a cap? Or is that what you're discouraging.

 

[Roff]

It doesn't matter. I'm working out a transfer function, and while the RL series combo is much more intuitive, I've got to use capacitors.

:roll:


So, can I obtain the frequency(w) from the RL series combo, and then use that to find an equivalent impedance with a cap? Or is that what you're discouraging.

Is this only at a single frequency? I feel like we're playing 20 questions. :evil:

 

[falmeshon]

magnitude of series RL impedance = sqrt (R^2 + (w*L)^2)
magnitude of series RC impedance = sqrt (R^2+(1/(w*C))^2)

where R is the resistance (stays the same in both circuits),
w is 2*pi*frequency of operation
L is inductance
C is capacitance

equate equations and solve for C

this is only valid when dealing with time varying signals and only equates the magnitude. The phases are not equated using this method, i.e you will not achieve the same thing by using capacitors.

 

[mstechca]

Believe it or not, the impedance of an RL series network is not equal to the sum of the individual impedances.

Why not?

 

[Russlk]

Mstecha: The reason is that the R and L are at 90 degrees phase angle, you have to use trigonometry.

It is not possible to get the same impedance from an R-C and R-L circuit, but if the signal is a single frequency and sinusoidal, the same magnitude of impedance can be obtained.

 

[Roff]

Mstechca, if you calculate or measure the impedance of a perfect reactance (inductor or capacitor), and call that X, and you put a resistor R in series, you can calculate the series impedance as

Z=sqrt(R^2+X^2)

or in plain English, Z is the square root of the sum of the squares of the resistance and the reactance.
This is only true for a series RL or RC circuit. It does not hold true if you have a series RLC circuit. It gets a little more complicated.