Simple Circuit

  • #1
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
 

Answers and Replies

  • #2
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Are any resistances in series or parallel?
 
  • #3
phinds
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Redraw the circuit, do current loops, show your work. Then we can help.
 
  • #4
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Break up the circuit into manageable bits. Think about which rules or laws you should be using.
 
  • #7
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Looks like a good start to me so far
 
  • #8
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
phinds
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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
Well with kvl it should add up to 0?
 
  • #12
309ly4j.jpg


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

Original circuit:
2qkikd3.jpg


Redrawn circuit:
(From this link) http://i44.tinypic.com/2hf7tc1.jpg
(Image size too large to imbed here directly.)

Can you see the mistake? :wink:

[Examine which resisters are on which sides of the circuit.]
 
  • #14
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!!

***Bangs head off wall****
 
  • #15
phinds
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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
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
phinds
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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
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
Curious3141
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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
Thanks curious, that certainty is a simpler approach
 

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