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enter image description here

If you read the second paragraph, it claims that the resistor and capacitor are in series, but they are both connected to same positive potential and both lead to ground, so isn't this a parallel configuration?

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    \$\begingroup\$ It's both series and parallel. \$\endgroup\$
    – Matt Young
    Oct 20, 2015 at 19:24
  • \$\begingroup\$ I tend to think that is a parallel connection. If it were series, then iC = iR \$\endgroup\$
    – bud
    Oct 20, 2015 at 19:40
  • \$\begingroup\$ @budder Wrong!! Applying KCL at the top node of the circuit iC+iR=0 \$\endgroup\$
    – GR Tech
    Oct 20, 2015 at 20:01
  • \$\begingroup\$ Both series and parallel. They may be considered as being in series once the dc source is disconnected as indicated by the first sentence. \$\endgroup\$
    – Tut
    Oct 20, 2015 at 20:16
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    \$\begingroup\$ When the source is suddenly removed it goes from a steady state parallel configuration to a temporary series connection from ground, where the capacitor is the new source of voltage, with a path to ground through the resistor. \$\endgroup\$
    – KyranF
    Oct 20, 2015 at 21:46

3 Answers 3

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Two dipoles are in series when the same current (same electrons!) are flowing through them.

They are in parallel if they are subjected to the same voltage (they are connected to the same two nodes at their terminals).

So yes, they are both in series and in parallel. It's not a problem --- it's like to ask if a runner in a party of two is the last one or the second one ;-).

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  • \$\begingroup\$ Rmano - I agree to the 1st and 2nd sentence, but not to your conclusion (3rd sentence). That means: They are NOT both in series and parallel, because they are not "subjected" to the same (external !!) voltage. They are to be considered as a series connection only. A parallel connection always requires a current splitting in a common node which here is not the case. When an external voltage would be applied at the upper node, both parts would be in parallel. \$\endgroup\$
    – LvW
    Oct 27, 2015 at 8:20
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It doesn't really make sense to consider this circuit a parallel configuration since there is no input at the node between the resistor and capacitor. It is simply a charged capacitor discharging into a resistor.

Note that:

$$i_R = -i_C$$

When there is only one current path, it is a series circuit.

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If you look at the diagram, you will see that one resistor terminal is connected to one capacitor terminal, and the the other to the other.

That means 100% of the current flowing in the resistor goes into the capacitor, and vice versa. This is the very definition of a series circuit.

Incidentally, one of those nodes is connected to ground. And the other is labelled +v. This doesn't change the connectivity.

If you want a meaningful parallel circuit, then you need a third current path (not just a high impedance meter) connected between the v node and ground, for instance a current source, another resistor, anything that will take a current. Then you can get different currents flowing in the R and the C, and the parallel description becomes worthwhile.

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    \$\begingroup\$ this is just wrong. \$\endgroup\$ Oct 20, 2015 at 20:47
  • \$\begingroup\$ agreed, wrong. all of it \$\endgroup\$
    – KyranF
    Oct 20, 2015 at 21:12

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