# Why resistance increase in a series connection and decrease in a parallel connection?

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.

• 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.
– eps
Commented Apr 1, 2023 at 18:07
• 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? Commented Apr 2, 2023 at 8:11
• Related: physics.stackexchange.com/a/671225/247642 (with the calculation based on simple version of Drude model) Commented Apr 3, 2023 at 8:50
• 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. Commented Apr 3, 2023 at 14:58

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.

• For the sponge test to give a fair view better use a wet (or dry) sponge twice. Commented Apr 1, 2023 at 14:43
• 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. Commented Apr 2, 2023 at 8:35

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.

• It was my pleasure. Commented Apr 3, 2023 at 4:04

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.

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.

• You are welcome. Commented Apr 2, 2023 at 13:55

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.

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.

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.

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.

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.

• You are welcome. Thanks to @ZaellixA for turning my touchscreen input into nicer formulas :) Commented Apr 2, 2023 at 8:49

When electrons flow into a resistor they are scattered, bouncing around much like in a pinball machine (see "Drude model": https://solidstate.quantumtinkerer.tudelft.nl/3_drude_model/). As they bounce around (a) they lose some of their potential energy (reduction of voltage), and as they are bouncing around (b) this means the resistor has an accumulation of negative charge which creates an electric force that resists further electrons from entering the resistor until some of the electrons exit on the positive side. (b) explains why resistors limit current.

If you place two resistors on the same path (in series), each has an accumulation of charge that produces electric forces pushing back against electrons that want to leave the negative pole. If you remove one of the resistors, then there's less total electric force and therefore a higher current can flow. If you add resistors in series, there's more total electric force pushing back.

Now in regards to simple circuits that contain only conducting wires and resistors, consider the following:

• the current (I) is the same at every point along a path in a circuit (a path goes from the negative terminal of a voltage source to the positive terminal of the same voltage source). Why? Consider what would happen if this were not true. The number of electrons would accumulate in an unbounded manner in one portion of the path. This would eventually result in a static discharge.
• at a particular point in a circuit all the electrons have the same potential energy (voltage V)
• the voltage (potential energy of the electrons) along a path is unchanged unless it is there is a resistor to reduce it
• the potential energy at the positive pole is 0
• you cannot know the current (I) of a particular path unless you know the total resistance of the path
• you cannot know the voltage drops of points along the path unless you know the current of the path

If there are two different paths from the negative pole to the positive pole (ie a parallel circuit), each path is treated independently because there is a virtually unlimited supply of electrons available to travel between the poles(1). The total current of the circuit will be the sum of the currents of each path. If the current in one path is huge (eg. there is no resistance on that path) then this means the current arriving at the positive pole is huge, so as far as the whole circuit is concerned, the effective resistance is almost 0, regardless of how large the resistance of the other path is.

If you combine series and parallel paths, then you need to use the bullet points I gave above to figure out what's going on. You need to put R, I, and V labels on all your known and unknown quantities, as shown in the diagram below.

We know that the current I2 in path 2 = V2/R2. We know that the current I3 in path 3 = V2/R3. We know that adding these two currents gives us the total current seen at point C on the circuit: V2/R2 + V2/R3 = V2/RP = I1, where RP is the effective resistance between points B and C. So, the effective resistance is the inverse of (1/R2 + 1/R3.)

Therefore the total resistance of the circuit is R1 + inv(1/R2 + 1/R3), and so current I1 = V1/(R1 + inv(1/R2 + 1/R3)).

Once we know I1, we can calculate currents I2 and I3, and the voltage drops produced by each of the resistors. We don't know these until we have the total current on the circuit.

Let's say you have a circuit with two voltage sources, and two paths across a particular resistor R1. In one path the current flows one way, in the other path it flows the opposite way. Electrons enter both end of the resistor, this creates a stronger force repelling electrons from entering either end, hence total current is reduced. Smaller current means each electron bounces around less as it makes its way through the resistor, hence losing less of its potential energy, hence the voltage drop in the resistor is smaller. The voltage drop in the resistor is abs(I1 - I2)R1. Hence Kirchoff's laws when you are calculating voltage drops around different paths.

(1) There are a lot of electrons in matter. One AA battery has around 2 x 10^22 electrons. As Fenyman noted, if two persons stood at arm's length from each other and each person had 1% more electrons than protons, the force of repulsion between the two people would be enough to lift a "weight" equal to that of the entire earth.