Parallel and Series-Parallel Resistance

Date: 05/24/2000 at 21:36:00
From: John Denning
Subject: Physic/Electrical Question

I am having problems calculating resistance in series circuits, 
parallel circuits, and series parallel circuits. I know the equation 
for parallel circuits is

    R_total = 1/r1 + 1/r2 + 1/r3 + ...

I know that they are inverted but I am having trouble.

For example: What is the total resistance in a circuit with 3 Ohm,
25 Ohm, 45 Ohm, and 67 Ohm resistors in parallel?

Thank you,
John Denning

Date: 05/25/2000 at 13:12:14
From: Doctor TWE
Subject: Re: Physic/Electrical Question

Hi John - thanks for writing to Dr. Math.

Using the 'reciprocal rule' for parallel resistance, you would take 
the reciprocal (1/x) of each resistor, add them up, and then take the 
reciprocal of the total.

For example, suppose we have a parallel circuit with resistors of
30 Ohms, 60 Ohms, 20 Ohms and 10 Ohms. Then the total resistance is:

     1     1    1    1    1
     -- = -- + -- + -- + --
     Rt   30   60   20   10

           2    1    3    6
        = -- + -- + -- + --
          60   60   60   60

          12   1
        = -- = -
          60   5

Thus

           1
     Rt = --- = 5 Ohms
          1/5

Of course, if you're using a calculator you don't have to work with 
the fractions themselves. On most scientific calculators, you could 
enter it like this: (Note that [1/x] is the reciprocal key and may be 
labeled [x^-1] on some models.)

     30 [1/x] [+] 60 [1/x] [+] 20 [1/x] [+] 10 [1/x] [=] [1/x]

On some calculators, you'll need to press the [=] key again at the 
end. Try my example on your calculator and see if you get 5. If not, 
look in your owner's manual to see how to use the reciprocal function.

An alternate way to solve for total resistance in parallel is the 
'product-over-sum rule'. This rule is often easier to use if you don't 
have a calculator with the reciprocal function handy. For two 
resistors, the rule states:

          R1 * R2     <- their product
     Rt = -------     <- over
          R1 + R2     <- their sum

If, for example, we had a 30-Ohm resistor in parallel with a 60-Ohm 
resistor, the total resistance would be:

          30 * 60
     Rt = -------
          30 + 60

          1800
        = ---- = 20 Ohms
           90

This agrees with the reciprocal rule's answer. (Check it!)

You have to be careful, however, when extending the rule to more than 
2 resistors. For 3 resistors, the rule becomes:

              R1 * R2 * R3
     Rt = ---------------------
          R1*R2 + R1*R3 + R2*R3

The 'product' part is simple, but the 'sum' part becomes the sum of 
the products of every 2-resistor combination. It gets even worse with 
more resistors.

Of course, you can use the 'product-over-sum' rule to find the 
parallel resistance of R1 and R2, then use it again to find the 
parallel resistance of that with R3, etc.

In the last part of your question, you asked about series-parallel 
circuits. Sometimes, a series-parallel circuit can be broken into 
series parts and parallel parts. In this case, each time you simplify, 
use the appropriate rule. For example, in the following circuit:

   

let R1 = 30 Ohms, R2 = 10 Ohms, R3 = 40 Ohms and R4 = 20 Ohms.

First, we can combine R1 and R2, which are in series. Call that Rx:

     Rx = R1 + R2 = 30 + 10 = 40 Ohms. (Series rule - add R's up)

   

Now we can combine Rx and R3, which are in parallel. Call that Ry:

     1/Ry = 1/Rx + 1/R3    (Parallel - use Reciprocal rule)
          = 1/40 + 1/40 
          = 2/40 = 1/20

    so Ry = 1/(1/20) = 20 Ohms.

   

Now we have a series circuit with Ry and R4. So our total resistance 
is the sum of these values:

     Rt = Ry + R4 = 20 + 20 = 40 Ohms

If you can't break the circuit up into series and parallel parts, you 
have to use Kirchoff's laws and Thevenin's theorems to produce a set 
of simultaneous equations for the voltages and currents, then use 
algebra to solve the system of equations. This can get very messy.

I hope this helps. If you have any more questions, write back.

- Doctor TWE, The Math Forum
  http://mathforum.org/dr.math/