# Simple Circuit

1. Dec 11, 2013

### DunceKirchhoff

http://i43.tinypic.com/2qkikd3.jpg

How do I tackle the above question? Ive got a few formulas in my head that dont help. Im not asking for answer or the working just a quick kick in the right direction.

Thanks

2. Dec 11, 2013

### willem2

Are any resistances in series or parallel?

3. Dec 11, 2013

### phinds

Redraw the circuit, do current loops, show your work. Then we can help.

4. Dec 11, 2013

Break up the circuit into manageable bits. Think about which rules or laws you should be using.

5. Dec 11, 2013

### DunceKirchhoff

Am I right so far

6. Dec 11, 2013

### DunceKirchhoff

7. Dec 11, 2013

Looks like a good start to me so far

8. Dec 11, 2013

### DunceKirchhoff

Well what I can abstract from that is-

4i1+10+3(i1+i2)
= 17i1+3i2

6i2+2+3(i1+i2)
=11i2+3i1

17i1+3i2=20
3i1+11i2=20

It doesnt figure, must have added up something wrong :P

9. Dec 11, 2013

### phinds

I don't really get how you got those equations, but they are both wrong and not even consistent with each other.

How much should the voltage around a loop add up to?

10. Dec 11, 2013

### DunceKirchhoff

Well with kvl it should add up to 0?

11. Dec 11, 2013

### phinds

So why is there not a zero on one side of your equations?

12. Dec 11, 2013

### DunceKirchhoff

I was using this as an example. I know there is an extra battery in it, but I have solved circuits in the past using this working with one battery.

What do you make of it?

13. Dec 11, 2013

### collinsmark

I think the first mistake was done in the process of redrawing of the circuit.

Original circuit:

Redrawn circuit:
(Image size too large to imbed here directly.)

Can you see the mistake?

[Examine which resisters are on which sides of the circuit.]

14. Dec 11, 2013

### DunceKirchhoff

Hi Colin

I placed the resistors on opposite sides, although giving me a more sensible answer,it was as they say on Catchphrase 'Its good but not right!!

15. Dec 11, 2013

### phinds

When doing loops, it is best to do independent loops to the extent possible. I see now what you did but I never use that method, as I find it cumbersome. That is, I think the way you have chosen your loops is awkward. BUT ... in any case, the sum around the loop should be zero. I ask again: why is there no zero in your equations?

16. Dec 11, 2013

### DunceKirchhoff

It has no zero in it because the method out the book does not show it. Yes I understand the sum of the voltage around the circuit adds up to zero, but look at the given example there is no use of this. I'm learning on my own so I use whatever method thats in front of me. I been learning physics for like 2 weeks :p

17. Dec 11, 2013

### phinds

Sorry, I seem to have implied that I didn't already understand what you just said, but again, I suggest that you drop that method and stick to simple loops that are as independent as possible and DO use the "sum to zero" around the loops,.

18. Dec 11, 2013

### Curious3141

Drawing current/voltage loops is a simple approach, but I would probably look for a more intuitive approach in this problem, which takes less than 5 minutes to get an exact answer. I prefer to simplify the resistor network wherever possible (and it's easy to do here). The circuit can be redrawn as the 20V cell in series with two resistances, one 3ohm, the other (10+6)||(2+4)ohm = 16||6ohm. The "||" notation refers to resistors in parallel. The first question is easy to answer after this. The other two questions can then be worked out by figuring out the current in each arm of the circuit (remember that the current is inversely proportional to the resistance in that arm as a proportion of the total effective resistance). Then just use V = IR to work out the p.d. across the resistor they asked for.

BTW, there are some rounding errors in the provided answers.

19. Dec 12, 2013

### DunceKirchhoff

Thanks curious, that certainty is a simpler approach