# Computing rlc circuit in parallel

So, if a resistor, inductor and capacitor in series sum algebraically, how do they work in parallel?

Is it $\dfrac{1}{\dfrac{1}{R}+\dfrac{1}{\dfrac{1}{L}+\dfrac{1}{C}}}$ ?

Joe

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No: capacitors in series don't sum algebraically; if you work with impedances it works, but if you work with capacitance you compute the equal of resistors parallel. If you have different components, you work with complex numbers, so you have to use impedances. –  clabacchio Apr 2 '12 at 14:41
Sorry, I knew this but misspoke. –  Joe Stavitsky Apr 2 '12 at 15:38
You can correct at any time :) –  clabacchio Apr 2 '12 at 15:42

Resistors and capacitors don't add algebraically. If your resistor is 1$\Omega$ and the magnitude of the capacitor's impedance is 1$\Omega$ their series circuit won't give you 2$\Omega$, but 1.4$\Omega$. That's because you add vectors which are at 90° with each other, which is represented by multiplying with $j$. The 1.4 is $\sqrt{2}$, which you get if you vectorially add two perpendicular vectors with modulus 1. So you'll have this factor $j$ in your equations.
Also, the impedance of capacitors and inductors is frequency dependent, it's not just $C$ or $L$. In the equation below this frequency dependence shows as the factor $\omega$.

$Z = \dfrac{1}{\dfrac{1}{R} + \dfrac{1}{Z_L} + \dfrac{1}{Z_C}} = \dfrac{1}{\dfrac{1}{R} + \dfrac{1}{j \cdot \omega \cdot L} + j \cdot \omega \cdot C} = \dfrac{1}{\dfrac{1}{R} + j \left(\omega \cdot C - \dfrac{1}{\omega \cdot L} \right)}$

Note the negative term between the brackets. This means that the imaginary part can become zero, namely when

$\omega = \sqrt{\dfrac{1}{L C}}$

In that case the parallel circuit becomes purely resistive with

$Z = R$

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Where does the 1.4 come from? If you don't use complex numbers, there is a high risk of misunderstanding –  clabacchio Apr 2 '12 at 15:02

For a series circuit, the impedances add. For a resistor, the impedance is $R$. For an inductor the impedance is $j\omega{}L$. For a capacitor the impedance is $1/j\omega{}C$.

So the impedance $Z$ of a series RLC is given by

$R + j\omega{}L + 1/j\omega{}C$.

For a parallel circuit, the admittances add. Admittance is like the complex version of conductance and is usually denoted by the symbol $Y$. Where we have a complex Ohm's law, $V=IZ$, we can also express this in terms of admittance as $I=VY$ . For a resistor, the admittance is $1/R$. For an inductor, the admittance is $1/j\omega{}L$. For a capacitor, the admittance is $j\omega{}C$.

So the admittance $Y$ of a parallel RLC circuit is given by

$Y=1/R + 1/j\omega{}L + j\omega{}C$.

But admittance is also just the inverse of impedance ($Z = 1/Y$), so we can simply invert this last formula to get the result given by StevenVH in his answer.

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another great one, thx –  Joe Stavitsky Apr 2 '12 at 15:49