# More than 2 resistors in parallel

Most people know the formula for the total resistance of parallel resistors:

$\dfrac{1}{R_t} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + {}...{} + \dfrac{1}{R_n}$

If there are only 2 resistors, that can be easily rearranged to solve for Rt:

${R_t} = \dfrac{(R_1 \cdot R_2)}{(R_1 + R_2)}$

Is there a safe way to do that for n resistors?

• Try math.stackexchange.com? – Kaz Nov 5 '12 at 1:54

Of course there is, but it does not look pretty. Make the divisors equal, and add the terms. for three resistors you get

$\dfrac{R2 \cdot R3}{R1 \cdot R2 \cdot R3} + \dfrac{R1 \cdot R3}{R1 \cdot R2 \cdot R3} + \dfrac{R1 \cdot R2}{R1 \cdot R2 \cdot R3}$

$= \dfrac{(R2 \cdot R3) + (R1 \cdot R3) + (R1 \cdot R2)}{R1 \cdot R2 \cdot R3}$

now do the $\dfrac{1}{n}$ and you get:

$\dfrac{R1 \cdot R2 \cdot R3}{(R2 \cdot R3) + (R1 \cdot R3) + (R1 \cdot R2)}$

The top line is easy, it is the product (multiplication) of all resistors. The bottom line is the sum of the products of all leave-one-out combinations. For two that reduces to the pretty formula:

$\dfrac{(R1 \cdot R2)}{(R1+R2)}$

• I approved a review here, please double check it since it is your post @wouterVanOoijen. It would also be nice if someone did some mathjax love here. – Kortuk Nov 4 '12 at 19:22
• Did someone say MathJax? – Adam Lawrence Nov 5 '12 at 0:10

Yes, it is, but the easiest and most compact representation is the

$${1\over R_t} = \sum_{i=1}^n {1\over R_i} = {1\over R_1} + \dots + {1\over R_n}$$

you already mentioned. To see the pattern you can do the math yourself for 3 resistors:

$${1\over R_t} = {1\over R_1} + {1\over R_2} + {1\over R_3}$$

Do the first addition using the usual method:

$${1\over R_t} = {R_1 + R_2\over R_1R_2} + {1\over R_3}$$

Then do the second addition, using the same method:

$${1\over R_t} = {R_1R_2 + R_2R_3 + R_1R_3\over R_1R_2R_3}$$

Solved for $R_t$:

$${R_t} = {R_1R_2R_3\over R_1R_2 + R_2R_3 + R_1R_3}$$

As you can see, the denominator is the sum of the products of "each with each" resistor value, which is more complicated to write than just

$${1\over R_t} = {1\over R_1} + {1\over R_2} + {1\over R_3}$$

When I teach introductory circuit classes, I always emphasize the notion of duality which, when mastered, can give you deep insight into many fundamental "rules" of circuit analysis.

The idea is that if you know the answer for, say, a series circuit, you can take the dual of the result and get a correct answer for a seemingly very different problem.

So, here is short list of circuit duals:

Voltage - Current

Resistance - Conductance

Inductance - Capacitance

Series - Parallel

Thevenin - Norton

There are others but these will do most of the time.

Ohm's law is usually written as:

$V = I R$

To take the dual, replace all the variables in the above equation with their duals:

The dual of Ohm's Law:

$I = VG$

where $G = \dfrac{1}{R}$

Recall that for resistors in series, resistances add, so that the equivalent resistance is just the sum.

Consider the dual of this, conductances in parallel.

From the principle of duality, parallel conductances add just as series resistances. So, if you have 3 conductances in parallel, the equivalent conductance is:

$G_{eq} = G_1 + G_2 + G_3$

Now, convert back to resistance:

$R_{eq} = \dfrac{1}{G_{eq}} = \dfrac{1}{G_1 + G_2 + G_3} = \dfrac{1}{\frac{1}{R_1}+\frac{1}{R_2} +\frac{1}{R_3}}$

In other words, the equivalent resistance of $n$ parallel resistors is the reciprocal of the sum of the reciprocals.

This is the origin of your first formula.

Actually, if you're doing a real problem with a bunch of resistors. What you would do is look for the "few" smallest resistors and compute their resistance. In other words, if you have 10 roughly equal resistors then the total resistance is about 1/10. If you have 3 small and 7 large, you can guess it is between 1/3 to 1/10, but if the 3 smalls are really small then it should be around 1/3 + delta. You should develop a sense of value when you see a resistance network. Guestimate is your friend :)