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I know the current is the same in a series branch. But does it make a difference, in the operation and purpose of the circuit, if elements arrangement order changed in a series connection?

Say a circuit to filter out AC signal using an AC source with DC offset, a resistor and a capacitor. What difference there is if the arrangement of the elements is changed?

In short words using an analogy: Does the current run like a human and it reacts to each element it sees first according to what the element orders it to do and it doesn't know what the future, upcoming circuits' elements ,holds for it?

EDIT:

First output using an arrangement: Circuit configuration-A Second output using another arrangement: Circuit configuration-B

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  • \$\begingroup\$ Consider your example of a resistor + capacitor creating a low-pass filter. If you reverse the order, you have created a high-pass filter. The same currents flow, but since the order is reversed the voltage produced can change. \$\endgroup\$
    – jbord39
    Aug 2, 2016 at 16:31
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    \$\begingroup\$ But the filters are not serial circuits. Their output is parallel to either component. \$\endgroup\$
    – Eugene Sh.
    Aug 2, 2016 at 16:34
  • \$\begingroup\$ That is true, but unless you are able to take a differential signal off the first component the order usually controls whether the serial circuit has a low-pass or high-pass behavior. Even though the same current flows either way, and the same voltages are induced in each component; depending on how you reference the voltages the order usually matters. \$\endgroup\$
    – jbord39
    Aug 2, 2016 at 16:42
  • \$\begingroup\$ @Khaled: Without a schematic your question is open to misinterpretation. There is a schematic button on the editor toolbar and it's easy to use. Double-click the component to edit its properties. \$\endgroup\$
    – Transistor
    Aug 2, 2016 at 17:03
  • \$\begingroup\$ @transistor I uploaded an example schematic and sorry for any inconvenience \$\endgroup\$ Aug 2, 2016 at 17:59

2 Answers 2

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I think the error is that you think you have a simple series circuit. This may have been true until you added the measurement device. Once you did that you created series-parallel circuits.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Series C and parallel R. (b) Series R and parallel C.

Figure 1a should indicate a little more clearly that C1 is in series with the signal while R1 is in parallel with the load. Figure 1b is the converse.

schematic

simulate this circuit

Figure 2. In these cases the components are truly in series. It will make no difference to the output signal which order R and C are in.

In the case shown in Figure 2 the effect on the output signal will be identical.

Does the current run like a human and it reacts to each element it sees first according to what the element orders it to do and it doesn't know what the future, upcoming circuits' elements ,holds for it?

You're starting to get into EM-wave theory here and it gets complex and not all that helpful in circuit analysis. Current is more like an incompressible fluid in a pipe being pumped around a circuit. All the water moves simultaneously at a rate determined by the circuit resistance. In the electrical circuit the current moves everywhere at speeds approaching the speed of light although the individual electrons move rather more slowly. In the water analogy the pressure is felt all around the circuit although a particular molecule of water may take minutes to travel around the circuit.

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  • \$\begingroup\$ Thank you very much for the detailed answer as this was exactly my confusion and now I get it :) \$\endgroup\$ Aug 2, 2016 at 18:44
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If we are talking about a pure (spherical) serial circuit (in a vacuum), the order of the components doesn't matter. But if we consider some simple appliance, which is exposed to external world, like, say, a heater with a switch, we can look at the two cases:

schematic

simulate this circuit – Schematic created using CircuitLab

Suppose the R is the heater (or any other load) with exposed body/contacts. Then, the circuit on the left will have these contacts "hot" (I.e. having voltage relative to the ground) even when the switch is open, so one can get electrocuted by touching the exposed part even when off. The circuit on the right doesn't have this problem as R will have zero potential when the switch is off.

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  • \$\begingroup\$ Thank you for your answer I understood. But this is explains that difference is in arrangement is true speaking of conductivity, as in the left picture I can get electrocuted since I will be the connection to ground. But speaking about circuits output according to elements's order , since for example a resistor has an effect and a capacitor has another effect and so on.. How does changing these elements' arrangement affect the output? \$\endgroup\$ Aug 2, 2016 at 18:01
  • \$\begingroup\$ Once you touch it, you become a part of a circuit, so basically it is not the original circuit anymore. This is why in the first sentence I said that if we look at pure serial circuits, without intervention from outside it wouldn't matter. \$\endgroup\$
    – Eugene Sh.
    Aug 2, 2016 at 18:03
  • \$\begingroup\$ With some arrangements the nodes may have totally different voltages depending on the order but the total voltage over the whole string will remain constant. The current will also remain constant unless you have branches or parasitic magnetic or capacitive effects. \$\endgroup\$
    – KalleMP
    Aug 2, 2016 at 23:10
  • \$\begingroup\$ I'd like to clarify that, in terms of "touching" (becoming part of the circuit that leads to ground) the switch contacts, there's little difference between these two circuits. In both, touching the "hinged" side flows current through you to ground (in the right circuit, the current wouldn't be resisted by the load, making it worse ... both cases are bad). I believe @EugeneSh. is saying that the the load's contacts are "hot" in the left circuit. In the right circuit, with the switch open, touching either load terminal is fine (as long as you're grounded). In the left circuit, not so much. \$\endgroup\$ Feb 14, 2018 at 21:26

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