Solving Ohm's Law Problem: Understanding Potential Difference in a Circuit"

[maxpayne_lhp]

Well, I'm quite idle in Lunar New year's holidays so I decide to do some exercises on Physics, I got stuck when solving the problem, then I tried with the suggestion on my textbook, but still, I don't understand the problem, please help me out! btw, sorry, something has gotten on LaTex or I'm too bad at it, so I just place the raw syntax, sorry!
Well, we have the figure:
http://img.photobucket.com/albums/v649/maxpayne_lhp2/5.jpg
And also, we have:
$$\emf=48V, r(Internam resistance)=0\omega$$
$$R_1=2\onega$$
$$R_2=8\onega\]$$
$$R_3=6\onega$$
$$R_4=16\onega$$
Well, they told me to find out the value of $$U_{MN}$$ (Well, in my native scientific system taugh in school,$$U_{MN}$$stands for the potential diferece between M and N.
Well, obviously, I found that
$$U_{MN} = U_{MA}+U_{AN}$$
$$=U_{AN}-U_{AM}$$
Then, I tended to account for the Intensity of the current thru AB via the formula:
$$I=\frac{\emf}{R_{AB}}$$
Then, I found out that I=8 (A)
But soon, I saw something wrong and looked up in the suggestion corner, they told me:
As r=0. we have
$$U_{MN}=\emf.$$ Let the current flows thru AMB and ANB from A to B, then:
$$U_{AN}=\frac{U_{AB}}{R_2+R_4} . R_2 = \frac{\emf}{R_2+R_4} . R_2= 16V$$
Do the same thing with $$U_{AM}$$
Though, I don't really understand what they mean, so please help me out! What did I do wrong?
Thanks for your time and help!

 

[Andrew Mason]

There is no resistance between M and N so there is no potential difference. M and N are at the same potential.

To find the potential difference across R1 and R2, work out resistance for R1 and R2 in parallel: $1/R_{1,2}=1/R_1 + 1/R_2$ and for R3 and R4 in parallel: $1/R_{3,4}=1/R_3 + 1/R_4$. Add $R_{total}=R_{1,2}+R_{3,4}$. Then work out the total current using $I = V/R_{total}$. That will allow you to work out the potential from A to M and from M to B.

AM

 

[maxpayne_lhp]

There is no resistance between M and N so there is no potential difference. M and N are at the same potential.

Ohh, sorry, I'm so clumsy, I didn't mean to draw a straight line connecting M and N, there was nothing connecting them at all. yes, they meant to cross the resistances.
And that's helped. Thanks!
BTW, I got difficulty when I need to go down for the next line, what's the syntax in LaTex for it? If you don't have time, please send me a link so that I can refer. Thanks a lot!

 

[xanthym]

Then the currents I_13 thru R_1 & R_3 and I_24 thru R_2 & R_4 are given by:

$$:(1): \ \ \ \ I_{13} = \frac {E} {R_1 + R_3}$$

$$:(2): \ \ \ \ I_{24} = \frac {E} {R_2 + R_4}$$

from which you can determine all required voltage drops. The overall equivalent resistance R will then be:

$$:(3): \ \ \ \ R = \frac {E} {I_{13} + I_{24}} = \frac {1} { \frac {1} {R_1 + R_3} + \frac {1} {R_2 + R_4} }$$


~~

 

[maxpayne_lhp]

Oh thanks, I got it, it's $$U_{MN}=4V$$
Thanks a lot!

 

[maxpayne_lhp]

Excuse me, still, I don't know what

$$:(3): \ \ \ \ R = \frac {E} {I_{13} + I_{24}} = \frac {1} { \frac {1} {R_1 + R_3} + \frac {1} {R_2 + R_4} }$$

for? Please explain for me.
Thanks!
btw, if we figure the another way, as thru R3 and R4, is it the same?
Thanks

 

[xanthym]

The equivalent resistance R is the value of a SINGLE resistor that can REPLACE ALL the resistors R_1, R_2, R_3, and R_4 between your battery. In other words, this resistor R will draw the same current that ALL your other resistors draw TOGETHER when connected to a battery. It's useful to know R in terms of the OVERALL operation of ALL the resistors. However, you still need to use the individual resistor values to determine voltage drops between the individual resistors inside your circuit -- just like you were doing.

 

[maxpayne_lhp]

Well, okay, Thanks!
Thanks for all of your time and help!