I am having a problem regarding calculating the impedance of the two circuits below:


simulate this circuit – Schematic created using CircuitLab

Everywhere I search I always get a description of what to do if they are connected in series or in parallell, however no one I have found so far explains how to use both parallell and series to make an Impedance function using the jw method. Would be extremly happy if someone explains the best method of getting the jw impedance function and also what does R1 // L1 stand for in these cases. Why is R1 // L1 often refered to as (R1 * L1) / (R1 + L1) and not (1/R1 + 1/L1) Like regular parallell functions.

The values on the different parts in the schematics are just examples, I would like to know how to make the formula using the signs like R1, C1 and so on.

Thanks in advance

  • \$\begingroup\$ R//L means in parallel and better to write \$R//X_L\$ latter part is missing an inversion in "regular" \$\endgroup\$ Commented Mar 11, 2018 at 19:53
  • \$\begingroup\$ Your first diagram can be found on wikipedia. en.wikipedia.org/wiki/RLC_circuit under 'other circuits'. \$\endgroup\$
    – Oldfart
    Commented Mar 11, 2018 at 20:11

3 Answers 3

I always get a description of what to do if they are connected in series or in parallell

That's ultimately all there is. You keep applying parallel and series combinations until there is only a single impedance left. For example, in your first circuit, R and L are obviously in series. Solve for that. Now compute that result in parallel with C. In the second circuit, you can likewise combine R2 and C1 in series, then the result of that parallel with R1, then the result of that in series with L.

The "//" operator means in parallel with. The arbitrary case of multiple impedances in parallel is the reciprocal of the sum of the reciprocals. R1 // R2 // ... Rn = 1 / (1/R1 + 1/R2 + ... 1/Rn).

Note that for just two impedances, this can be reduced to (R1 * R2)/(R1 + R2). Your second equation for parallel impedances is wrong because you forgot to take the reciprocal of the resulting sum.

Note that the math above works not just for resistances, but also arbitrary impedances. In the general case, these are complex numbers. If all the impedances are resistances, then all the values are real numbers, making the computation easier. However, the algebra is the same, except that it has to be carried out on complex numbers in the case of arbitrary impedances.


First you identify the concerned impedance is across which two points. Let's take the first circuit. The total load impedance seen by the voltage source will be across the voltage source, ie., across the capacitor C. Mark the two nodes. Now between the nodes of the capacitor, R and L are series, as same current flows through them. so you add them to get one equivalent impedance, say Zi, where R = R and XL = jwL. Now Zi is in parallel with the capacitor, as voltage across Zi and C are same. So the net impedance will be (Zi || Xc) = (Zi*Xc)/(Zi+Xc), where Xc = -1/jwC. The laws of serial and parallel, which you apply with "resistor only circuits", apply even to inductances and capacitances in the circuit too. They form what we call "linear circuits".


Well, for the first circuit we can write:


And for the second one:



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