Why is PD zero across the 3Ω resistor and why count all battery internal resistances?

[aznking1]

Homework Statement

I have two questions. Both questions are in the picture below. Dont mind my workings lol

http://img163.imageshack.us/img163/5680/scanpic0001qe.jpg

Ans to the 1st qn: C

The potential difference across the 3ohm resistor is zero. Thus, it can be considered that the 4ohm and 2 ohm resistor is in parallel with the 8ohm and 4 ohm resistor. So the effective resistance between A and B is

(1/(4+2) +1/(8+4))^-1 = 4ohms

But my question is, why is the p.d. across 3 ohm resistor zero?

Ans to the 2nd qn: C

The effective e.m.f is 1.5v and effective internal resistance is (1/3 + 1/3+1/3)^-1 = 1ohms
E=I(R+r)
1.5=I(10 + 1)
I=0.14A

And my question is, how does the circuit work? Why do we have the consider all of the internal resistance of the batteries?

 

[Infinitum]

Question (1) is a typical wheatstone bridge. This bridge has potential difference in the middle resistor equal to zero if the other resistors have the relation

$$\frac{R_1}{R_3} = \frac{R_2}{R_4}$$

Where resistors are taken R1, R3, R4, R2 clockwise from A in your diagram. You can prove this using Kirchoff's laws.

For question (2)
Every resistor is an opposition to the flow of current. Battery's internal resistance is not an exception to this. Electrons try to cross the battery and are obstructed within. So you do have to consider its resistance. To make it simpler, you can even assume it as a separate resistor just beside the battery, as it makes no difference to the problem.

 

[azizlwl]

Question (2)
As you see all the internal resistors which normally taken as a series resistor to the emf.
All are connected at one end. The other ends all have the same potential.
So you can connect all in parallel have a single battery.