how to work out total resistance of a parallel circuit

[georgeo5664]

hi i am new to learning about electrics i have just started an electricians course and have a question. The question is : Two heating elements of 70 ohms & 30 ohms are connected in parallel the total resistance of this circuit would be......

i have tryed all differnt web sites i have worked it out using (proof sum) r1 x r2 210
(r1 + r2) = 100 = 2.1ohms

but the formula that i was given to work with was as follows example:

1 = 1 1
- + - -
Rt R1 R2

sorry if this is not the best explained description but i am new and trying

 

[georgeo5664]

thank you for all that has took time to help me but can you work the formula out without the use of a calculator many thanks

 

[i=p/u]

number crunching for calculators

 

[rob@etc]

georgeo5664 welcome to the wonderful world of resistors in series and parallel!!
when you have 2 resistors in parallel the simple formula is THE PRODUCT OF THEM DIVIDED BY THE SUM OF THEM so
the product (70) multiplied by (30) divided by the sum (70) plus (30)
so 70 x 30 = 2100 / 70 + 30 = 100
so 2100 /100
= 21 ohms
when its more than 2 resistors in parallel then your formula applies. Hope that helps

 

[georgeo5664]

georgeo5664 welcome to the wonderful world of resistors in series and parallel!!
when you have 2 resistors in parallel the simple formula is THE PRODUCT OF THEM DIVIDED BY THE SUM OF THEM so
the product (70) multiplied by (30) divided by the sum (70) plus (30)
so 70 x 30 = 2100 / 70 + 30 = 100
so 2100 /100
= 21 ohms
when its more than 2 resistors in parallel then your formula applies. Hope that helps

Thanks alot Rob@tec so i can not use the formula to work it out unless it has 3 or more resistors is that right

 

[rob@etc]

your formula will still work on 2 resistors but as a rule of thumb yes 3 or more, dont forget lowest common denominator and dont forget to flip the calculation because its 1
Rt

 

[georgeo5664]

right i have mastered the first one i just cant get my head around how to do using the formula anyh help would be appreciated

 

[Richard Burns]

Lenny's post #14 is very clear but I show the calculation in older style mathematical notation in this picture

and you can also write this as


For Robs formula with two resistors in parallel this can be written as

Hope this helps

 

[ackbarthestar]

Lenny's post #14 is very clear but I show the calculation in older style mathematical notation in this picture
View attachment 8995
and you can also write this as
View attachment 8994

For Robs formula with two resistors in parallel this can be written as View attachment 8997

Hope this helps

Excellent couple of posts here.

Just to add the basic proof

Basic circuit formula
1/ ........ I = Vs / Rt

For parallel circuits the voltage across each resistor will be the same, so the only parameter that will vary will be the current flow
So according to Kirrchof's law for parallel circuit.... the current entering a node is equal to the current leaving a node etc etc
2/ Is = I1 + I2 + I3

Substitute Vs / R for each of the current flows in each resistor and we have...
3/.........Is = Vs / R1 + Vs / R2 + Vs / R3

Now substitute Vs / Rt for the total current drawn from the supply (Is)

4/........Vs / Rt = Vs / R1 + Vs / R2 + Vs / R3

Now comes the maths bit

We can simplify this equation by dividing through by Vs:

5/...... Vs / Rt * Vs = Vs / R1 * Vs + Vs / R2 * Vs + Vs / R3 * Vs

Now Vs / Vs = 1 so:

6/.......1 / Rt = 1 / R1 + 1/ R2 + 1 / R3.

Ok?

A further simplification of this and used as shorthand for 2 resistors in a circuit.

A bit more maths

Can 1 / R1 + 1 /R2 be represented in another way? Well Yes it can.

If we want to add these two quantities together we need to first find the common denominator

Which is : R1 * R2 ...... Think about this example 1 /4 * 1 / 3 = (3 + 4) / (4 * 3)

So
1/....... 1 / R1 equiv R2 / R1*R2
2/.........1 / R2 equiv R1 / R1*R2

So adding 1/R1 + 1/R2 gives us

3/........ 1/Rt = R1 / R1*R2 + R2 / R1*R2

Now combine the 2 together, since they now both share the same common denominator, and we have:

4/ 1/Rt = R1 + R2 / R1*R2
But Rt is not the same as 1/Rt so we need to invert the whole lot which gives us:

5/ Rt = R1*R2 / R1 + R2 (product over the sum rule)

Try both ways with small whole numbers such as 1/3 + 1/4

and for those doing 2391

The simplest way of calculating resistors in parallel is:

1/ Convert the fraction into a decimal faction 1/2 = 0.5, 1/3 = 0.33, 1/4 = 0.25 1/ 299 = 0.003344

2/ Add all the decimal fractions together

3/ Invert the answer (1/x or x^-1 on your calculator)

I hope this helps, (also available for high-wire circus juggling performances)

 

[Knobhead]

Ye GOD's just how much have I forgotten!

This thread is bringing back suff I've not used since collage 30+ years ago.

Thanks Georgio for starting it
:welcome: