# Need help with series-parallel resitor networks

If I have a parallel resistor network with different values for each resistor, how do I calculate the total resistance? Sorry I dont have enough rep to post pics so I'll try and explain.

Lets say I have a matrix of 9 resistors total in a 3x3x3 grid with 3 different values for the 9 resistors, say 10 ohm, 20 ohm and 30 ohm. So that makes 3 sets of 10, 20 and 30 ohms.

I need to calculate the total resistance of the 9 resistor network.

• "9 resistors total in a 3x3x3 grid". It's possible to have a circuit in which no components are connected in series or in parallel. Would you post a schematic of your circuit? – Alfred Centauri Jul 11 '12 at 23:12
• I would but I get the error "you don't have enough REP to post pics" The schematic is on page 29 of 'Electronics Demystified' if that helps. – acidblue Jul 11 '12 at 23:15

You add the conductances (reciprocal of resistance) and then convert back to resistance. For example, lets say we have 500, 1500, and 2500 ohm resistors in parallel. The total resistance is:

1 / ((1/500) + (1/1500) + (1/2500)) = 326.1 ohms.

For 2 resistors in parallel you can also use (R1 * R2) / (R1 + R2) e.g. for 500 and 700 ohms:

(500 * 700) / (500 + 700) = 291.7 ohms.

For your example of the 10, 20 and 30 ohms in a 3 x 3 network, you can either work out each row then add the results, or multiply each value by three and calculate directly:

    |
10 20 30
10 20 30
10 20 30
|


1 / ((1/30) + (1/60) + (1/90)) = 16.3636... ohms.

If the example is like this (in/out at top and bottom of matrix):

    |
10 10 10
20 20 20
30 30 30
|


then the result is 20 ohms (row 1 is 3.33... row 2 is 6.66... row three is 10 - added together is 20) which agrees with the answer you have.

• OK so in my case could I do the following: 1/ ((1/10)+(1/20)+(1/30)) = 1/(0.1833) = 5.455 Then since there are 3 sets 3*5.455 = 16.366 ?? – acidblue Jul 11 '12 at 23:12
• Yes, either that or the other way shown above in the (edited for your example) answer. – Oli Glaser Jul 11 '12 at 23:16
• AHA! I thought so, But that is so weird cause the book i got the schematic from says the answer is 20 ohms, maybe they are rounding up?? – acidblue Jul 11 '12 at 23:19
• I doubt it is an intentional round up, possibly a mistake - I'd have to see the example to confirm. You can upload the image somewhere and post the link, then we can add it to the question. I was assuming 3 rows of 10,20,30 for the example. – Oli Glaser Jul 11 '12 at 23:22
• See edited answer for arrangement that adds up to 20 ohms - is this the circuit in your book? – Oli Glaser Jul 11 '12 at 23:26

First I'll show that series resistances add. Let's take one branch of 10 Ω + 20 Ω + 30 Ω resistors. The current through each of them is the same, let's give it a name: $I$. Then according to Ohm's Law we have the following voltages across the resp. resistors:

$(I \times 10 \Omega)$ across the $10 \Omega$ resistor,
$(I \times 20 \Omega)$ across the $20 \Omega$ resistor,
$(I \times 30 \Omega)$ across the $30 \Omega$ resistor,

for a total voltage of

$V = (I \times 10 \Omega) + (I \times 20 \Omega) + (I \times 30 \Omega) = I \times (10 \Omega + 20 \Omega + 30 \Omega)$

so that

$R_{total} = \dfrac{V}{I} = (10 \Omega + 20 \Omega + 30 \Omega) = 60 \Omega$

The total resistance of resistors in series is the sum of their resistances.

So we can simplify to three 60 Ω resistors in parallel. This time we won't have the same currents but the same voltage across each resistor. Then the currents through the branches are:

$\dfrac{V}{60 \Omega}$ through the first $60 \Omega$ resistor,
$\dfrac{V}{60 \Omega}$ through the second $60 \Omega$ resistor,
$\dfrac{V}{60 \Omega}$ through the third $60 \Omega$ resistor,

for a total current of

$I = \dfrac{V}{60 \Omega} + \dfrac{V}{60 \Omega} + \dfrac{V}{60 \Omega} = V \times \left(\dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega} \right)$

so that

$\dfrac{I}{V} = \dfrac{1}{R} = \dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega}$

and thus

$R = \dfrac{1}{\dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega}} = \dfrac{1}{\left(\dfrac{3}{60 \Omega}\right)} = \dfrac{60 \Omega}{3} = 20 \Omega$

The equivalent resistance of $N$ resistors of $R$ Ω each is $R/N$ Ω.

In practice you'll often have two different resistors in parallel, then you have to use the equation I mentioned earlier:

$R = \dfrac{1}{\dfrac{1}{R1} + \dfrac{1}{R2}} = \dfrac{1}{\left(\dfrac{R1 + R2}{R1 \times R2}\right)} = \dfrac{R1 \times R2}{R1 + R2}$