Computing rlc circuit in parallel

Question

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}}}\$ ?

Many thanks in advance

Joe

Answer 1

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 \$

Answer 2

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.