1. Now, normally I wouldn't like to ask for help, but this is really bugging me, and I'm surely missing some fundamental idea. (Plus, I'm not really sure about the hw help policy on this board, though just looking through a few threads I did notice there were some appeals for help.) What are the (a) magnitude and (b) direction (up - type "1" or down - type "0") of current i in the figure below, where all resistances are 6.5 Ω and all batteries are ideal and have an emf of 10.0 V? (Hint:This can be answered using only mental calculation.)
Is this simply a matter of a really long loop rule and junction rules? I thought otherwise because it said it could easily be solved using only mental calculations. So maybe there is a concept I should know?  2.

3. As far as I can see, you would need to use nodal analysis. Select nodes to use as common ground nodes and then find the potential at all other nodes. you can calculate voltages across resistances by the difference in potentials at each end, and this also gives you the direction of current as well as the magnitude. Using KCL, you know that the sum of currents entering a node must be equal to the sum of currents exiting the node.

You could first try and simplify the problem by finding equivalent resistances i.e....series: R[total] = R + R...and parrallel: 1/R[total] = 1/R + 1/R...

I'm not sure about the mental calculation of the current, but maybe the numbers are easy enough to work with that you could do it in your head and label the circuit as you go?  4. Yes, you can do it in your head if you are patient enough.

Start by simplifying the diagram. For example, did you notice that the four resistors in the lower right corner (southeast) can all be redrawn as a single resistor?

Study the diagram and you will find other opportunities for simplification. Also note that some emf cells act in series, others in parallel. Can you simplify further?

Keith, take it!
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Â*  5. This is definatelt partly a 'trick' question, the reference to 'can only be done mentally' is definately a reference to the simplification, looking at it you can see several simplifications, then further ones as you work out voltages etc. Im sitting in work. gonna give this a shot to keep me occupied. 8)  6. on second thought the question asks for current and direction through each resistor yes? So simplification is banned unless its to help with calculations elsewhere.   7. Originally Posted by harvestein
on second thought the question asks for current and direction through each resistor yes? So simplification is banned unless its to help with calculations elsewhere. No, only through the resistor marked with the i (on the middle left side).  8. yeah i just noticed that when i printed it out a few times. Here goes..... 8)

Cheers  9. ive simplified it a lot, but my nodal analysis is well rusty. Thats what happens when you bum out of uni and work a dead end jub for a year. wikipedia time. 8)  10. Â*
Aw c'mon Keith. Give it a try yourself. It's a cute problem. Keep simplifying.

Here's another hint: the six resistors in the northwest and north central can all be reduced to single resistor.

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Â*  11. Ohh, I will definitely give try it myself. I have to learn this stuff for tests (specifically the AP). Right now though, I've got to get through the rest of "The Sun Also Rises" for lit. I'll give it a go after that. One issue though, I don't know that I have ever been taught how to combine emfs, so does anyone have any insight into that?  12. For cells in series, the total EMF is the sum of the EMF's of the cells. (Also, total ESR is equal to the sum of individual ESR's...not needed here, but useful to know). For cells in parrallel, the total EMF is simply the maximum EMF of the individual cells. For example, If you have a 5V, and 10V cell in parallel, the EMF will be 10V...not factoring in the ESR...which the question tells you to leave out anyway.

The rest is just simplifying resistor networks, and a bit of nodal analysis, which is basically Kirchoff's circuit laws.  13. Here is the hint to that problem:

You need to find a solvable loop, that is, one involving a single unknown current, the one requested. If your path choice forces you to introduce another (unknown) current into the loop equation, you are sunk.

ps. I meant to do it already, but I have just been inundated with extraneous busy work.  14. Look at the lower left hand side...start by simplifying the four resistances there into a single value, where the current i will pass through. Then look at the potentials either side of this resistance...i.e. the node above and below it...then find the potential difference across the resistance, and work out the current.  15. I got it. This yields the answer quickly:

8V/5R

I got that from:

(1/R+1/R)^-1 + R + R -->5R/2

V/(5R/2) = 2V/5R

4*(2V/5R) = 8V/5R  16. How did you get 5R/2 from (1/R+1/R)^-1 + R + R ?

Hints:
-the current goes up
-you have left out a resistor.  17. I think I know why it says, "This can be answered using only mental calculation". All the batteries are connected by wires without resistors. This makes the problem far easier than you might expect.  18. Originally Posted by bit4bit
How did you get 5R/2 from (1/R+1/R)^-1 + R + R ?

Hints:
-the current goes up
-you have left out a resistor.
(1/R+1/R)^-1 = R/2 + R + R

I told you though, I got it right, 8V/5R was correct.  19. Originally Posted by Keith
I told you though, I got it right, 8V/5R was correct.
Yep that's the answer I got. 4(10 volts) across the 2.5(6.5 ohms).  Bookmarks
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