- #1

- 581

- 20

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Helly123
- Start date

- #1

- 581

- 20

- #2

BvU

Science Advisor

Homework Helper

- 14,448

- 3,737

If you leave out the 3 and 6 the current is already more than 1 A, so your friend must be wrong

- #3

- #4

BvU

Science Advisor

Homework Helper

- 14,448

- 3,737

I'm not a betting man, nor a teacher. You have to convince yourself !

- #5

BvU

Science Advisor

Homework Helper

- 14,448

- 3,737

Le me ask: what do you get when you leave out the 6 and the 9 ?

- #6

- #7

BvU

Science Advisor

Homework Helper

- 14,448

- 3,737

( leaving out the 6 and 9 leaves only the 3 ! )

- #8

cnh1995

Homework Helper

Gold Member

- 3,444

- 1,144

They are not.6 ohm and 3 ohm are series = b

Can you see anything special about the 6 ohm resistor?

- #9

- 581

- 20

Wouldn't the current go to both branches? Im still really not good at it. But why only 3?

( leaving out the 6 and 9 leaves only the 3 ! )

I = 9.6/3 = 3.2 A?

- #10

- 581

- 20

Yes, i think so. Maybe because the current from 1st branch go to 6ohm, and the current from 2nd branch go to 6 ohm too? So they're not series...?They are not.

Can you see anything special about the 6 ohm resistor?

If pararel then I = 9.6 / (18/11) = 5.8 A?

- #11

stockzahn

Homework Helper

Gold Member

- 498

- 137

- #12

BvU

Science Advisor

Homework Helper

- 14,448

- 3,737

The word is parallel.Yes, i think so. Maybe because the current from 1st branch go to 6ohm, and the current from 2nd branch go to 6 ohm too? So they're not series...?

If pararel then I = 9.6 / (18/11) = 5.8 A?

Hwat is 11/18 ?

How far are you now ? Can you see the 6 ohm is short-circuited ? There is a wire from one end to the other. That wire makes the potential at both ends equal -- no current through the resistor. You can leave it out of the diagram without anything changeing. Then things look simple again ...

- #13

- 581

- 20

Oh parallel.. thanksThe word is parallel.

Hwat is 11/18 ?

How far are you now ? Can you see the 6 ohm is short-circuited ? There is a wire from one end to the other. That wire makes the potential at both ends equal -- no current through the resistor. You can leave it out of the diagram without anything changeing. Then things look simple again ...

3ohm and 9ohm are parallel, the total resistance is 9/4

I = 9.6 / (9/4)

I = 4.2 A is it?

- #14

BvU

Science Advisor

Homework Helper

- 14,448

- 3,737

As in #4: you have to convince yourself !

(But I like this one a lot !)

(But I like this one a lot !)

- #15

- #16

- #17

gneill

Mentor

- 20,925

- 2,867

What options are available?To be honest, i think i still get wrong. My answer is not among the options..

- #18

cnh1995

Homework Helper

Gold Member

- 3,444

- 1,144

To be honest, i think i still get wrong. My answer is not among the options..

You have chopped off some decimal places.I = 9.6 / (9/4)

I = 4.2 A is it?

What are the options?

- #19

epenguin

Homework Helper

Gold Member

- 3,873

- 897

Is an option 5⅓ A?

- #20

mpresic2

- #21

BvU

Science Advisor

Homework Helper

- 14,448

- 3,737

So 3.2 A for the branch with the 3 ##\Omega## only.Wouldn't the current go to both branches? Im still really not good at it. But why only 3?

I = 9.6/3 = 3.2 A?

That would be the 3.2 we had already plus 1 A for the 9 ##\Omega## resistor ? But how much is 9.6 V / 9 ##\Omega## ?Ok.. we can conclude it as 4.2 A. Thank you

- #22

- 581

- 20

- #23

BvU

Science Advisor

Homework Helper

- 14,448

- 3,737

Perhaps you can see for yourself if you first erase the I and then add some V in your picture: both ends of R

And then your I

- #24

epenguin

Homework Helper

Gold Member

- 3,873

- 897

I don’t know whether this is the only reason, anyway this calculation is taking far too long! I think we have got that the 6Ω resistor has no potential difference across it, there is no current in it, it might as well not be there. So all you have we have is a single voltage source and two resistors in parallel. But that is still being found, and is often found by students, unnecessarily difficult. Let me just give my take and tips on why is simple resistive (and capacitive) steady current circuits are found more difficult than they need be.

They are found unnecessarily difficult, in my guess, because students are taught something about reciprocal resistances being the sum of other reciprocal resistances or something, which sounds arcane, mathematical, and calculational, distanced from the simple physics. Which is that current comes out of one pole of the source, divides into streams, which then reunite and the

You can just calculate each of the two currents in this problem separately. Then add them together for the total current.

Didacts have maybe tried to make this simpler by a formula, but I’d say experience shows they have made it more difficult for students.

When you have got used to adding up currents in parallel parts, which won’t take long, you may realise that parallel branches with the same potential difference across them are acting like a single conductor where the ability to conduct a current – the

I = GV

(You can call this a definition. Many people would call that equation Ohm’s Law. Really what Ohm’s law - a generalisation from experimental findings - says is that for conductors made of many materials, and over a wide range of conditions, G is a constant.)

Then the way currents in parallel add up mean that conductances in parallel add up. I.e. several conductors G

G = G

which hopefully has a self-evident quality about it.

Obviously from the equation above, conductance G is the reciprocal of resistance R. I guess dicacts have tried to not overburden students with too many terms and more or less eliminate ‘conductance’ from their terminology, but the term is suggestive so I think they have lost more than they gained.

When resistors are in series on the other hand, it is potential differences that add up, and the resistance concept is more useful as the total resistance of a series of resistors is the sum of resistances.

My

Have an eye for and use any symmetry in a circuit to simplify calculation. (I only got involved in homework help for circuits because I was seeing students making unnecessarily complicated calculations which could be simplified by symmetry.)

Use rational numbers like 5⅓ in preference to decimals. Using decimals will often be wiping out information. And decimals are boring. Rationals bear a trace of the original circuit problem. (These are my reasons - I won’t make exaggerated claims for accuracy. Just once in 250 times when making accurate calculation on complicated circuit you might gain an accuracy in the third significant figure by using rationals, and of course the input data is generally not that accurate, it’s just a point of principle.) Decimals are good only for the final answer, because measuring devices use decimals.

- #25

BvU

Science Advisor

Homework Helper

- 14,448

- 3,737

Beg to differ. I think Helly did better in # 17, apart from a small rounding error. Problem is that the right answer isn't among his/her options!I think I gave the right answer in #19

Share: