# Resistors in Parallel

I understand that when you put Resistors in parallel the resistance decreases, my question is why does this actually happen? What causes the Resistance to decrease? When Resistors are in serial, their Resistances are added, why when we put them in parallel do the Resistors resist less current? I don't understand how that's possible?

• Think about it like adding lanes to a highway - by adding more ways to get directly from A to B, it becomes easier to do it. In other words, the resistance decreases. – JustJeff Mar 12 '12 at 11:02
• +1 for being very similar to my water and pipes metaphor :) – Kaos Mar 12 '12 at 11:07

Instead of thinking of these things as "resistors", try thinking of them as "conductors". After all, that's what they do: conduct.

A resistor with resistance $R$ is a conductor with conductance $S=\dfrac{1}{R}$.

When you provide multiple conductors connecting one point to another, the conductances simply add. What could be more intuitive? When you provide an additional path for current to flow, more total current flows.

Conductors $S_1$ and $S_2$ in parallel have a total conductance of:

$S = S_1 + S_2$

If you want to express $S_1$ and $S_2$ as resistances $R_1$ and $R_2$, you get:

$S = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

And, if you want to express the total conductance S as a resistance R:

$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

$R = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}}$

Which is the usual expression for the total resistance of two resistors in parallel.

Trivia: the unit of conductance (i.e. inverse ohms) is sometimes called the "mho" ('Ohm' backwards), and is written with an upside-down Omega symbol: ℧. The official SI name for this unit is siemens ("S").

• It makes a lot more sense when you see where the actual equation comes from, thanks. – John Mar 12 '12 at 21:29
• +1 nice explanation with the use of S. And I'd never heard about mhO but Siemens is better :) – clabacchio Mar 13 '12 at 8:24
• This makes a lot of sense when you write it this way, until I go back to the fundamental idea that there are two paths, each of which with R resistance, why should a current passing through either one have less than R resistance?? I understand what you said and it makes sense that way, but somehow I'm not overcoming this other erroneous way of thinking about it. It's like if there are two paths through the woods and both have poison ivy, the fact that there are two doesn't mean there's now less poison ivy lol. – temporary_user_name Jun 11 '15 at 21:16
• @temporary_user_name (OK, really this is for future readers). The resistance of each individual path doesn't change. The effective resistance of the combination is less than the resistance of either individual path. – The Photon Dec 27 '18 at 19:04

The total resistance is given by Ohm's Law:

$R = \dfrac{V}{I}$

For two resistors in parallel the voltage stays the same, but current increases since it follows two paths. Suppose V = 1V, and each resistor is 1k$\Omega$. Then for one resistor the current will be 1mA. If there are two 1k$\Omega$ resistors there will be two 1mA paths, that's 2mA. Per Ohm's Law increased current means lower resistance:

$R = \dfrac{1V}{2mA} = 500\Omega$

If resistors are not equal the smaller resistor will carry the largest current. The other resistor(s) will add to this current, so the total current is always larger than that of the smallest resistor. Therefore the equivalent resistance is always smaller than the smallest resistor.

Each resistor still has the same resistance. By adding resistors in parallel, you are providing more conduits for current to flow; therefore the overall effective resistance of the parallel setup decrease. Think of resistors and wires as pipes - the larger the resistance, the narrower the pipe. Wires have negligible resistance, thus they are the widest pipes around. When you add resistors in series, the current still has only one "pipe" to flow through, but you are connecting narrow pipes one after another. Thus the overall resistance is much larger than before.

Multiple resistances in parallel each draw current and the effective resistance is the resistance that would draw the same current as the combined resistances do.

Consider the following version of Ohm's Law

• $R = \dfrac{V}{I}$

If you have a "black box" with two wires connected and are told that there is a resistor inside you could measure voltage applied and current drawn to determine the internal resistance.

If you apply 10V and 1 mA is drawn you conclude that $R = \dfrac{V}{I} = \dfrac{10}{0.001} = 10k \Omega$.

Now consider that there are TWO resistors inside and that they are in parallel. Again apply 10V and you will see that 2 mA (not 1 mA as before) is drawn. 1mA will flow through 1 resistor and 1 mA will flow through the other resistor.

Looking from outside you see $R = \dfrac{V}{I} = \dfrac{10}{0.002} = 5,000 \Omega$.
Because what is inside draws the same current as $5000 \Omega$ you know that inside there IS either $5000 \Omega$ or something of equivalent resistance. Clearly $2 \times 10,000 \Omega$ in parallel has $5000 \Omega$ resistance.

ie multiple resistances in parallel each draw current and the effective resistance is the resistance that would draw the same current as the combined resistances do.

$i_1 = \dfrac{V}{R_1}$

$i_2 = \dfrac{V}{R_2}$

$i_3 = \dfrac{V}{R_3}$

$i_{total} = i_1 + i_2 + i_3 = V \times \Big(\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\Big)$

$R_{effective} = \dfrac{V}{I_{total}}$

So $I_{total} = \dfrac{V}{R_{effective}} = V \times (\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3})$

so $R_{effective} = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}$

The effective resistance = the inverse of the sum of the inverse resistances.

eg 100 ohm + 200 ohm in parallel

$R_{effective} = \dfrac{1}{\dfrac{1}{100} + {1}{200}} = \dfrac{1}{0.01 + 0.05} = \dfrac{1}{0.015} = 66.666 \Omega$

When I was first introduced to electronics, my mentor (rather than tutor, as it was not within a school setting) he used water, pipes and basins as a metaphor to make it easier to understand. And for me, that worked really well (already at the age of 12).

Say you have a basin of water, with a pipe at the lower end, leaking some of the water out:

|~~~~~~~|
|_______===...


Now, it is easy to imagine that if you change the width of the pipe connected to the basin, it will affect the flow of water out of the basin.

A smaller pipe, less water, a big pipe, more water flowing out.

Now, we can extend this metaphor, saying that we don't need to replace the entire pipe, but if you just replace a section somewhere along the length of the pipe, it will affect the flow of water for the whole pipe. It won't matter if we have a really big pipe, if there's a tiny pipe at the far end, we will still just have a little flow of water.

This is still at a very basic level, but say if we connect more pipes to the basin (in parallell), we will get more flow of water out of the basin (less resistance).

We can take the metaphor further, but this should be far enough to answer the question...

• Water is a good metaphor, but gas is even better because electrons behave like an ideal gas in a conductor. It is also easier to explain, how a capacitor works. – Ariser Jan 20 '17 at 15:22

Unlike sawing a 2x4 in half, electronics is THEORY and the theory is proven through math. The physical resistance of each resistor does not change, only the amount of CURRENT going through each resistor changes. The previous explanation of the highway is a very good one. Think of crowded highway driving as a STRESS FACTOR. The stress factor, resistance to your well being decreases as the number of lanes increase and the resistance of the traffic decreases. All of this is proven through math. BTW... I REALLY LIKE THE WAY Ric/Russel Posted the formulas. Barry