Calculating total resistance

bacnka

This is going to sound like such a stupid question, but I'm a bit of a pickle:

Below is a circuit



I'm being asked to work out

[a] total resistance
[b] the cell current
[c] the p.d, current and power delivered to each resistor
[d] the electrical power delivered by the cell

right, when starting [a] would my question be something like:

1/4 + 1/12 + 3 = R[total]

I'm just a *bit* stuck

Thanks for the help anyway! (I know it's probably the most stupid question in the world)

John

StatusX

No, look at the units in that equation. First find the equivalent resistance for the two parellel resistors, then add this to the third.

bacnka

I'm seriously too lost to work that out :(

comwiz72

Don't forget the equations for this:

Resistors in series: R(T) = R1 + R2 + R3 and so on...

Resistors in parallel: 1/R(T) = 1/R1 +1/R2 + 1/R3 and so on...

where R(T) is the total resistance.

But you must work out the series and parallel bits separately. First, forget about the final resistor and consider the two that are in parallel. Then work out the total for the whole circuit.

Hope this is clear and it helps!

bacnka

I seriously don't know how to do it though

adding the two parallel ones together

like:

1/4 + 1/12

= 3/12 + 1/12

= 1/3

I guess that isn't the right thing to do?

comwiz72

No your right so far...

what you have calculated so far is 1/R(T) which you have found to be 1/3. So, you need R(T). Can you work it out from here?

bacnka

it would be 3 + 1/3 ?

So it would be 3.33333333... ?

comwiz72

no no...

You still haven't worked out the actual resistance for the parallel part yet. You worked out that 1/R(T) = 1/3

Now, to get from 1/R(T) to R(T) you turn over the fraction right? Well, you know that doing one thing to one side of an equation means you have to do it to the other side, so by the same principle...

1/3 becomes ...?

Once you get this, you can redraw your circuit diagram, and replace the two parallel resistors with a single one with resistance equal to the answer to "1/3 becomes ...?".

This means that the calculation becomes a simple one with just two resistors in series, which is simple to calculate.

Hope this helps!

Tom Mattson

bacnka, I know that you said that you are too lost to heed the suggestions by StatusX and comwiz72, but you are going to have to get un-lost in a hurry if you expect to pass this course. The point they are making is so basic that you will not make it without "getting it".

First of all: Think, don't guess.

These numbers have units on them. You aren't just adding 3 to 1/3, you are adding 3 OHMS to 1/3 INVERSE OHMS. You cannot add two quantities that have different units.