# (Fairly Basic) Voltage Divider Problem

Discussion in 'Homework Help' started by ||Steve||, Dec 18, 2009.

1. ### ||Steve|| Thread Starter New Member

Dec 18, 2009
4
0
Hi I'm new here (found the site after searching for an age for help)
I'm having some issues with the above rule that I think someone might be able to help with.
I'm working on a question similar to the one attached. I understand the basic "equation" for Voltage Divider (my version is in the picture) but I can't figure out how to find the voltage at point a. Can't remember any rule I learnt for it and I've a test tomorrow. Any help at all would be appreciated (and duly awarded with +rep or something).

BTW As for attempts, I have spent an hour or two searching lecture notes to no avail. I can do similar questions that don't involve 2 resistors on one "branch" easily enough.

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Take a look here at the AAC ebook material on voltage divider.

hgmjr

3. ### ||Steve|| Thread Starter New Member

Dec 18, 2009
4
0
Apologies, I'm from Ireland and used to seeing V for voltage. It won't happen again ha. Thanks again the tutorials/guides are excellent.

4. ### Fraser_Integration Member

Nov 28, 2009
142
5
Those two branches are in parallel, so you can say that they both have 10V across them.

Then you can apply the potential divider equation to the 2.2k and 3.3k series branch.

5. ### zgozvrm Member

Oct 24, 2009
115
2
1) Calculate the total resistance (series-parallel)
2) Find the total current.
3) Find the current through each branch.
4) Find the voltage across each resistor.

The voltage at point "A" is the voltage measured from point "A" to ground which, in this case, is the same as the voltage across the 3.3KΩ resistor

6. ### Rick Martin Active Member

Jun 14, 2009
31
2
Use voltage divider formula which for this is:

[3.3k/(3.3k + 2.2k)] * 10v

= 6v at a guess.

As others have said because it is a parallel network the voltage across the 2 resistors will be 10v.

You dont need to work out the total resistance of the cicuit UNLESS you want to calculate some form of current with in the circuit.

7. ### ||Steve|| Thread Starter New Member

Dec 18, 2009
4
0
Thanks for the help guy's. I'd say I have it "sussed" now, now to just sit this test...

8. ### JMD Member

Dec 9, 2009
96
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There's no need to all that - its a waste of time. Its simple ratio calculation.

I remember it by "Which do you want to find the voltage across?" - in this case its the 3.3k. Then the 3.3 needs to be the numerator, and the denumerator is the two resistors in series. This means:

10*(3.3/(3.3 + 2.2)) = 6v

In this case, the 2k resistor can be ignored - you only need to look at that one, in case you need to find the total impedance or total current.

9. ### zgozvrm Member

Oct 24, 2009
115
2
Not to someone who's still learning. You've got to understand the basics first. Plus, if you forget the voltage divider formula, you can always fall back to good ol' Ohm's Law.

10. ### Rick Martin Active Member

Jun 14, 2009
31
2
This is true but he did ask for the voltage divider formula, hence why all the other calculations were not required for this perticular problem.