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I recently posed this question to someone. I always thought that the answer was simply no because of Kirchhoff's voltage law... However, their response was... "no, the voltage is not altered due to the total resistance staying the same" ... does this have any merit? I guess I never thought about it that way, but I don't see why they'd be wrong in saying that. Would just like some clarification on this.

For example, if I had two sources in a series circuit that both produced 2V as well as two 5 Ohm resistors (also in series). No matter how I positioned the sources or resistors, there would always be a total of 4V produced and 4V dropped.

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  • \$\begingroup\$ What do you mean by "the order of any component"? \$\endgroup\$
    – Hearth
    Commented Feb 9, 2019 at 1:55
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    \$\begingroup\$ what do you mean by total voltage? ...... are you talking about power supply voltage placed across the series circuit? \$\endgroup\$
    – jsotola
    Commented Feb 9, 2019 at 1:55
  • \$\begingroup\$ @jsotola Go ahead and see my update. \$\endgroup\$
    – Snoop
    Commented Feb 9, 2019 at 1:57
  • \$\begingroup\$ @Hearth Go ahead and see my update. \$\endgroup\$
    – Snoop
    Commented Feb 9, 2019 at 1:57
  • \$\begingroup\$ Sorry about the confusion. \$\endgroup\$
    – Snoop
    Commented Feb 9, 2019 at 1:57

5 Answers 5

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To a first order no, but it somewhat depends on what do you mean by component and how are these connected in the series circuit.

  • Assuming all the components are two terminals and no parasitics to other parts of the circuit, then yes, KVL and KCL says that nothing would change. You need to add KCL as component voltages depend on their current (e.g., ohm's law).

  • If some components are non-linear (e.g, LEDs), but still two terminals and placed in the same order with respect to the circulating current, then yes, KVL and KCL says that nothing would change.

  • But, if the components are more than two terminals, even if there is no current through that extra terminal (e.g., FETs) then no, as the voltage with respect to that terminal can change the component's I(V) curve.

  • And, if there is current through that terminal or any other branch connected to the series circuit, then of course not.

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    \$\begingroup\$ If there are any devices with more than two terminals, the term "series" no longer makes sense. \$\endgroup\$
    – Hearth
    Commented Feb 9, 2019 at 4:31
  • \$\begingroup\$ @Hearth At that point it wouldn't be series anymore. \$\endgroup\$
    – Snoop
    Commented Feb 9, 2019 at 14:45
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    \$\begingroup\$ @Hearth you can always find people referring to any loop of connections as a “serial” connection, or putting a transistor or relay “in series” with the load. It’s an abuse of the term, but common enough to warrant accounting for. \$\endgroup\$ Commented Feb 9, 2019 at 14:51
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In a circuit with constant voltage sources and only resistive components, the order they are placed in to make the circuit will not affect the voltage drop across each of the elements will not be affected.

With a 10 V source, a 2 R, 3 R, and 5 R resistor will each experience a voltage drop of 2 V, 3 V, and 5V respectively.

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An alternative perspective comes from conservation of energy, which is more fundamental than KVL and KCL.

Conservation of energy implies that that power generated, must be equal to the power dissipated in a closed system. Since your circuit has the elements in series, the current is the same through each device. So by using passive sign convention with a common current, conservation of energy requires the total supply voltage to equal the total voltage drop summed across all the resisters, independent of ordering.

The fundamental idea for a closed system like your idealized circuit model, is that every watt of power that you put into a system (i.e., power generated), must come out as heat (i.e., power dissipated). This is something worth remembering for anyone who designs electrical or electromechanical systems.

Notes:

  • Passive sign convention on Wiki: https://en.wikipedia.org/wiki/Passive_sign_convention

  • Yes, this assumes that generated power is magically there, but the concept still applies in real world systems.

  • Total supply voltage is the algebraic sum of the supply voltage with respect to their polarity (+ - symbols).

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Even I think the voltage is not altered because there is no change in total resistance. According to your question you have 2 voltage sources of 2V and 2 resistors of resistance 5ohm. So the voltage remains constant until the resistance remains 10 ohms

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  • \$\begingroup\$ The resistance has nothing to do with it, though. Changing the resistance doesn't change the voltage at all. \$\endgroup\$
    – Hearth
    Commented Feb 9, 2019 at 15:00
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The response "the voltage is not altered due to the total resistance staying the same" is absolutely correct. The total resistance is what determines the current flow in your series circuit. And when you have your current, you can calculate the voltage at every component. And these voltages stay obviously the same, when the current stays the same, although the order of the components might be changed.

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  • \$\begingroup\$ This is circular reasoning. You're saying the resistance determines the current and the current determines the voltage, but the voltage and resistance determine the current, not just resistance. It's not really a cause-and-effect relationship in the manner you're claiming it to be. \$\endgroup\$
    – Hearth
    Commented Feb 9, 2019 at 15:03
  • \$\begingroup\$ Now you are looking at the individual components right at the beginnging - you can not do this in this scenario. The current flow is determined by the overall resistance (and, as you said, the voltage - but the complete voltage over the whole series circuit!). And this current then gives you the individual voltages. Obviously all this is linked, but this explanation gives the TO an easy picture of why the argument was correct. \$\endgroup\$
    – jusaca
    Commented Feb 11, 2019 at 7:00

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