Series & parallel

[thatricksta]

Ok physicists, hopefully I don't look like a fool here, but I'm trying to work out what is in series and what is in parallel for this circuit.

This is taken from the STAV 2009 trial exam, but I had to draw it my self because I don't have a copy on my PC (so excuse the poor paint job).

First of all, the question asks for the voltage across the terminals X-Y, if someone could explain this, that would be fantastic! The solutions don't provide a very good explanation.

Secondly, if anyone is willing to redraw this so it clearly displays series and parallel components that would be greatly appreciated.

Thanks!

[Lasercookie]

You didn't upload the picture...

[thatricksta]

Crap! Sorry LOL I got caught up in writing my paragraph =/

on this note, if anyone has the STAV 2009 exam could they please upload it i want some of the diagrams off it for my cheat sheet!

[Lasercookie]

That circuit is pretty much a voltage divider.

You can simplify the resistors down to:
1k in a parallel branch, with the 2k + 3k in the other branch.
50V gets supplied to each branch.

So your voltage divider is the 2k and 3k resistors, with a Vin of 50V.
X-Y is the voltage across the 3k resistor.

From there you can choose to use the ratio method (2:3) or the voltage divider equation.
I'm more comfortable with the voltage divider method.

[thatricksta]

That circuit is pretty much a voltage divider.

You can simplify the resistors down to:
1k in a parallel branch, with the 2k + 3k in the other branch.
50V gets supplied to each branch.

So your voltage divider is the 2k and 3k resistors, with a Vin of 50V.
X-Y is the voltage across the 3k resistor.

From there you can choose to use the ratio method (2:3) or the voltage divider equation.
I'm more comfortable with the voltage divider method.

So, the 1k is in series with the 2 and 3 in parallel?

Is this diagram a fair representation? (attached)

[Vincezor]

Yes. I'd even go as far as redrawing the circuit like that if I were to do that question.

[thatricksta]

Yes. I'd even go as far as redrawing the circuit like that if I were to do that question.

Thats exactly what I would do, hence asking I want it on my cheat sheet. Thanks!!

[Lasercookie]

Yeah, the tricky part (well for me anyway) with these simplifying diagrams is determining how it's all placed.

Q12 in 4.1 in the Heinemann book had this really odd circuit and it required you to simplify it down. I would take a shot at it. I reckon if you can do that simplify that diagram, you'll be right for the exam.

[thatricksta]

Yeah, the tricky part (well for me anyway) with these simplifying diagrams is determining how it's all placed.

Q12 in 4.1 in the Heinemann book had this really odd circuit and it required you to simplify it down. I would take a shot at it. I reckon if you can do that simplify that diagram, you'll be right for the exam.

would this be it??

[Lasercookie]

Not quite, the 30k and 10k resistors are in parallel, but they are in series with the 5k.

I set my diagrams out from top-to-down, that's just a personal thing really; I find it easier to think things through that way.

I also went and simplified the circuit down to the one resistor.

[thatricksta]

thank you!