# Equivalent RC circuit to a RRC circuit?

I'm in doubt about a situation that I've seen sometimes: imagine we have a resistor in parallel with a resistor and a capacitor in series. Since I don't know how to generate figures of circuits to post here, the situation can be described as: a single resistor on the right, and on the left a resistor and a capacitor in series.

If there was no capacitor, I know I could replace the resistors by an equivalent one. My doubt is, do this continues to be true in this case? I mean, can I replace this configuration by one capacitor with one resistor in series such that this resistor is equivalent to the other two? If we can, what's the argument beyond this?

-
electronics.stackexchange.com has an easy-to-use editor for circuit diagrams. You could flag your question for moving there. – RedGrittyBrick May 21 '13 at 19:19
Questions like this which are about simple circuit elements (resistor, capacitor, inductor) are generally considered to be on topic here. So this probably wouldn't be migrated. – David Z May 21 '13 at 19:31
Added a schematic of an example circuit like you've described. – Alfred Centauri May 21 '13 at 20:55

I mean, can I replace this configuration by one capacitor with one resistor in series such that this resistor is equivalent to the other two?

For a single resistor and capacitor in series, the real part of the impedance is independent of frequency, i.e., the real part acts like a resistor.

$Z_s = R_s + \frac{1}{j \omega C}$

However, for the circuit you describe, the real part varies with frequency. This is easily seen by noting that at zero frequency, the impedance is real and equal to the value of the parallel resistance. At "infinite" frequency, the impedance is real and equal to the parallel combination of the two resistances.

The equivalent impedance of the circuit you describe is:

$Z_{eq} = R_p || (R_s + \frac{1}{j \omega C})$

This can be expressed as the sum of a real part and an imaginary part but the real part involves the radian frequency $\omega$.

So, although we can write the above as the sum of a real (resistive) part and an imaginary (reactive) part, the real part acts like a frequency dependent resistor.

If your circuit were to be operated at a single frequency, then, in that limited context, the answer is yes, one can replace the two resistors with an equivalent resistor for that particular frequency.

-
Just let me note that $j$ is $\sqrt{-1}$. EE people prefer that over $i$ because $i$ is a current... – safkan Jan 11 '15 at 22:54
@safkan, indeed, that is why I carry a jPhone. – Alfred Centauri Jan 11 '15 at 22:55

This can be done, by extending the idea of resistance to the idea of impedance. You have to do this because the resistance of a capacitor depends on the frequency of the input voltage. If the input voltage is a pure DC signal with zero frequency, the capacitor will act like a resistor with infinite voltage.

Impedance is a complex number that acts like resistance. The impedance of a resistor is just the real number $R$ corresponding to it resistance. The impedance of a capacitor with capacitance $C$ is the complex number $Z_\mathrm{cap} = 1/i\omega C$, where $\omega$ is the frequency of the input voltage signal. (NB: electrical engineers like to use $j$ to represent the imaginary unit rather than $i$. The Wikipedia article uses that convention.)

Impedance adds using the same series and parallel addition rules that resistance follows. Because the result is a complex number describing real-valued physical quantities, you have to be careful about the interpretation. You can write the total impedance as $Z_\mathrm{total} = x + iy$ in Cartesian form, or as $Z_\mathrm{total} = r\exp(i\phi)$ in polar form. The magnitude of the impedance $|Z_\mathrm{total}| = x^2 + y^2 = r$ acts like a traditional resistance in that $I = V/|Z_\mathrm{total}|$. The phase shift $\phi = \arctan(y/x)$ is the phase by which the current lags the voltage.

-