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I was wondering that why does the equivalent resistance actually increase in a series connection of resistors and why does it actually decrease in a parallel connection of resistors?

I know that we can prove it by the formula $V/I = R$ but what's actually the reason behind it.

The current remains the same in series connection while the voltage changes and the opposite happens in a parallel connection so why don't the current change and voltage change just cancel out the effect of each other.

It would be helpful if you were to answer without the use of complex terminology.

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    $\begingroup$ A real life wire is basically a resistor and what do you do to have it carry more amps? Make it thicker, ie run a bunch of resistors in parallel. $\endgroup$
    – eps
    Apr 1, 2023 at 18:07
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    $\begingroup$ It is not clear what you mean exactly in your third paragraph, but if, either in the series or parallel scenario, only one of voltage or current changes, then how could the change(s)(there is only one) cancel out each other? $\endgroup$ Apr 2, 2023 at 8:11
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    $\begingroup$ Thank you for thanking me for my answer. If you find an answer useful, please upvote and/or accept. This will help others with the same question to find useful answers more easily. $\endgroup$
    – Roland
    Apr 2, 2023 at 11:46
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    $\begingroup$ Related: physics.stackexchange.com/a/671225/247642 (with the calculation based on simple version of Drude model) $\endgroup$
    – Roger V.
    Apr 3, 2023 at 8:50
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    $\begingroup$ In addition to V/I=R you also need Kirchhoff's laws (en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws) in order to derive the formulas for series/parallel connections. Combined, everything makes more sense. $\endgroup$
    – fraxinus
    Apr 3, 2023 at 14:58

9 Answers 9

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I was wondering that why does the equivalent resistance actually increase in a series connection of resistors and why does it actually decrease in a parallel connection of resistors?

Water analogy: wires are pipes, resistors are pipes with a sponge in it, water is electrical current.

To get through a sponge, the water will flow into holes which eventually end, and then into another hole that the first one connects to, and so on. The more distance it has to go, the more resistance it will see because it has to get through more of the holes along the path. Counteracting this is the surface area of the sponge presented to the flow. The larger the surface, the more "initial holes" you have to flow into, so the water can spread out and flow in parallel through more paths and thus more water gets out the other end in any given time.

Easy demonstration: take a normal household sponge, wet it, and then wring it out. Now hold it flat under your sink faucet and run the water. How long did it take to come out the back? Now wring it out and do it again, but this time hold it so the top edge is facing the water and it has to flow through the entire length of the sponge. How long did that take? Resistance.

So if you place two sponge sections in series, the water has to flow through both, one after the other. This is like when you hold the sponge the long way to the water. So the resistance is twice.

But if you put two sponge sections in parallel, then the water only has to flow through one thickness, and has twice as many holes to start from. This is like when you hold the sponge flat to the water. So the resistance is half.

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  • $\begingroup$ For the sponge test to give a fair view better use a wet (or dry) sponge twice. $\endgroup$ Apr 1, 2023 at 14:43
  • $\begingroup$ Thanks for answering my question! This analogy helped me to understand it better and from what I learned from all of the answers and replies is that the main factor here is the increase/decrease in "surface area" (or "paths") for current which leads the resistance increase in series and decrease in parallel. $\endgroup$ Apr 2, 2023 at 8:35
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The resistance of a resistor is given by $$R=\frac{\rho L}{A}$$ where $\rho$ is the resistivity, L is the resistor length and A is the cross-sectional area of the resistor.

For 2 identical resistors in series, the current passes through the same cross sectional area but the equivalent resistor is now twice as long:

$$R_1 + R_2 = \frac{2\rho L}{A}$$ For identical resistors in parallel, the cross-sectional Area doubles since the current is divided between the two:

$$\frac{1}{R_T} = \frac{1}{R} + \frac{1}{R}$$

$$\frac{1}{R_T} = \frac{A}{\rho L} + \frac{A}{\rho L}$$ $$ R_T = \frac{\rho L}{2A}$$

This effective doubling of resistive cross-sectional area lowers the resistance.

Resistors in series create a longer resistor. Resistors in parallel create a "wider" ( greater cross-section) resistor.

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    $\begingroup$ Thanks for answering my question! $\endgroup$ Apr 2, 2023 at 8:36
  • $\begingroup$ It was my pleasure. $\endgroup$ Apr 3, 2023 at 4:04
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Imagine a resistive device as a turnstile gate, and electrons as people trying to get through. A small gate represents lots of resistance that lets little current through for some voltage, while a large gate represents a small amount of resistance that lets lots of current through for some voltage. Putting another gate in the same wall in parallel always makes it easier to get to the other side, resulting in lower resistance and greater current for some fixed voltage. Those two gates may have different flow between them, however, if one is small and one is large.

On the other hand, putting two turnstiles in a series only makes it harder for people to get through, increasing the resistance and decreasing the current. Since the gates are one-after-another, anyone who goes through the first must also pass through the second at the same rate (or else electrons/people would "bunch up" somewhere), so these two gates must have identical current regardless of their size/resistance.

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  • $\begingroup$ Thanks for answering my question! $\endgroup$ Apr 2, 2023 at 8:35
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Resistance impedes current, just like a thin, rough pipe impedes water flow. Resistors in series reduce current like pipes in series reduce water flow; resistors in parallel increase current like pipes in parallel increase water flow.

Compared to one resistor, two equal resistors in series half the current, while two equal resistors in parallel double the current.

Ohm's law, $V = IR$ has been shown to satisfy the observed relationship between voltage and current for resistance.

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  • $\begingroup$ Thanks for answering my question! $\endgroup$ Apr 2, 2023 at 8:35
  • $\begingroup$ You are welcome. $\endgroup$
    – John Darby
    Apr 2, 2023 at 13:55
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In parallel circuit, to add more resistors, we need to add more "paths" to the circuit (a path for each resistor). As there are more than one path for current to flow, more current is flowing (even though the flow is resisted in each path) while keeping the voltage constant. This clearly shows that resistance is less in parallel circuit as compared to series circuit.

In a series circuit, first resistor will resist current flow and then the second resistor will resist current flow and so on. Thus, the amount of current has decreased keeping the voltage constant. So, at the end we can conclude that resistance in series circuit is more as compared to parallel circuit.

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Imagine two resistors in series, $R_1$ and $R_2$. Because of Kirchhoff's Current Law we know the current through $R_1$ is the same as the current through $R_2$ at all times. Say the current through both resistors is $I$.

By Ohm's Law, the voltage across $R_1$ is $V_1=IR_1$ and the voltage across $R_2$ is $V_2=IR_2$.

The voltage across both is $V_{total} = V_1+V_2 = IR_1 + IR_2 = I(R_1 + R_2)$ - the combination acts like a single resistor with a resistance of $R_{total} = \frac {V_{total}}{I_{total}} = R_1 + R_2$.


Conversely, imagine two resistors in parallel, $R_1$ and $R_2$. Because of Kirchhoff's Voltage Law we know the voltage across $R_1$ is the same as the voltage across $R_2$ at all times. Call the voltage $V$.

By Ohm's Law, the current through $R_1$ is $I_1 = \frac V{R_1}$ and the current through $R_2$ is $I_2 = \frac V{R_2}$.

The current through both is $I_{total} = I_1 + I_2 = \frac V{R_1}+\frac V{R_2} = V(\frac 1{R_1}+\frac 1{R_2})$

$=\frac V{\frac 1{\frac 1{R_1}+\frac 1{R_2}}}$

which is the same as a resistor with resistance $R_{total} = \frac {V_{total}}{I_{total}} = \frac 1{\frac 1{R_1}+\frac 1{R_2}}$

The latter set of equations are neater if you speak about conductance instead of resistance but I didn't want to introduce that.

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Okay, assume that you are a observer, standing near a road watching vehicles. If the road is wide many vehicles can go in same time or if road is narrow less vehicles can go through it. Now just imagine.... vehicles are electrons or positively charged particles (old concept) and road is the cross-section of conductor. Same applies in resistors they are just weak conductors. In series we increase length, now electrons have to go through a long way,more resistance,in parallel we increase cross section so more electrons can cross.

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Often, formulating the problem properly solves most of the problem. You probably mean to ask:

  1. Why does the resistance of a circuit increase when you add another resistor in series of the circuit?

  2. Why does the resistance of a circuit decrease when you add a resistor in parallel to the existing circuit?

The answer to 1. is obvious. Regarding to 2. you could rephrase as:

  1. Why does the conductivity of a circuit increase when you add another conductor in parallel to the circuit?

Resistance is just the inverse of conductivity. Now the answer to 2. is obvious too.

Do not see this as just a silly word-game. In computations of electrical networks the conductivity is used very often.

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    $\begingroup$ Thanks for answering my question! $\endgroup$ Apr 2, 2023 at 8:36
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Series:

$$ V_{1} = I R_{1} $$

$$ V_{2} = I R_{2} $$

$$ V = V_{1} + V_{2} $$

$$ R_{total} = \frac{V}{I} = \frac{V_{1}}{I} + \frac{V_{2}}{I} = R_{1} + R_{2} $$

Same for parallel, if you use $G = \frac{1}{R}$ instead, and replace later.

It‘s no magic.

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    $\begingroup$ Thanks for answering my question! $\endgroup$ Apr 2, 2023 at 8:36
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    $\begingroup$ You are welcome. Thanks to @ZaellixA for turning my touchscreen input into nicer formulas :) $\endgroup$
    – MS-SPO
    Apr 2, 2023 at 8:49

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