# Simplifying and identifying parallel circuits

1. Oct 7, 2011

### ashlynne_

I have not been able to solve these 3 questions. (attached)
I understand that parallel circuits can be identified by checking if there are branches at a particular point, but I am still unable to simplify these circuits and solve these questions.
Any help would be greatly appreciated!
Thanks! :)

Last edited: Oct 7, 2011
2. Oct 7, 2011

### Staff: Mentor

HI ashlynne_. Welcome to Physics Forums.

You'll have to provide some attempt to solve a problem before we can help. Why don't you pick one of your circuits and give us your thoughts on what's happening in it, and perhaps show an attempt at some calculations.

3. Oct 7, 2011

### ashlynne_

Okay..thanks.
for the first problem, this is what I was able to come up with:

I don't know how should I carry on from there.
Does the current actually split when it is between point X and the 4.8Ω resistor?

Last edited: Oct 7, 2011
4. Oct 7, 2011

### Staff: Mentor

That's a great start. No, the current doesn't split; the remaining components are all in series, so the current is the same through all of them (The wire leading to X carries no current because there's nowhere for it to go --- it's just a reference point where you might stick a voltmeter lead, for example).

If you were to find the current that the 8V battery will push through the resistors, you might then consider how to apply Ohm's Law to find the voltage drops that occur across each of them.

5. Oct 7, 2011

### ashlynne_

Okay,

I've worked out that the resistance across the 4.8Ω resistor is 3.10V,
and that through the 7.6Ω resistor is 4.90V.

So the current through the circuit is 0.645A, and I make use of this value to multiply with the total resistance through the circuit?

Last edited: Oct 7, 2011
6. Oct 7, 2011

### Staff: Mentor

Can you show your calculations in detail? Something seems to have gone awry somewhere.

Oh, by the way, voltage (potential difference) appears across components like resistors, while current passes though them.

7. Oct 7, 2011

### ashlynne_

Oh, okay!

This is what I did:

8. Oct 7, 2011

9. Oct 7, 2011

### ashlynne_

It's alright :)

Uhm, but how do I calculate the potential at X? Between which points do I look for the voltage drop?

10. Oct 7, 2011

### Staff: Mentor

Potential is always defined with respect to some reference point. In this case that reference point is indicated by the ground symbol at the bottom of the battery. One way to find the potential at point X is to find a path through the circuit that goes from the reference point to point X. Add up all the potential differences that occur along that path.

11. Oct 7, 2011

### ashlynne_

Oh, okay!
I get it now!

So in the case of the 2nd problem, the reference point would be point Z? & can I assume it to be at 0V?

12. Oct 7, 2011

### Staff: Mentor

You can choose any point that you wish, so long as you're consistent throughout the analysis. Some points will be more convenient than others (strategically placed) for the analysis.

When you are looking for the potential difference between two points, sometimes it's convenient to choose one of those points as the overall circuit reference point, sometimes it's not (the circuit reference point may be specified already, or the choice would make calculating other things more complicated). Whatever the choice of overall circuit reference point, the potential difference between two points, say X and Z, can be found by first finding the potential at each of those points with respect to the circuit ground, then taking the difference: E = X - Z.

For this circuit I personally would choose the negative terminal of the battery as the ground reference and then find the potentials at X and Z with respect to that.

13. Oct 7, 2011

### ashlynne_

Oh, I see. But how do you determine which part of the circuit would be more convenient as the ground reference?

14. Oct 7, 2011

### Staff: Mentor

It's one of those things that comes with experience. But in general, first look for the negative terminal of the power supply for the circuit if there's only one of them. If there are several, if they all tie together at one node, choose that. Failing that, choose a node that is a 'popular connection point' for several branches. The idea is to select a reference point that either has some overwhelming reason for being considered THE reference point, or is simply conveniently located for 'reaching' other nodes.

As you start writing equations to solve more complex circuits you will soon develop the intuition for choosing a reference point that makes the equations fewer and simpler

15. Oct 7, 2011

### ashlynne_

Okay!
I understand it now.
Thank you so much for your prompt replies and explanations! (: