resistance

Zidane

2011-09-20 6:21 am

圖片參考：http://farm7.static.flickr.com/6163/6162374083_b2525509f8_b.jpg

圖片參考：http://farm7.static.flickr.com/6179/6162919984_ffd631af20_b.jpg
1. need to find equivalent resistance across AB, how to obtain the graph on the right?
2. need to find equivalent resistance across AC, how to obtain the graph on the right?
3. need to find equivalent resistance across AB, how to obtain the graph on the right?
4. need to find equivalent resistance across BC, how to obtain the graph on the right?
5. need to find equivalent resistance across PQ, how to obtain the graph below?
6. What does the deflection of galvanometer imply?

thanks

更新1:

and also please answer the following post: http://hk.knowledge.yahoo.com/question/question?qid=7011091800650 (post it in the wrong area -.-) thanks

更新2:

for 1, Is it because the potential difference between the upper and lower point of the centre resistor is zero, so no current flow through? If yes, how can I show it mathematically?

更新3:

for 5, is it because the other 4 resistors are short-circuited, so no need to consider? If yes, how can I show it mathematically?

更新4:

and I still cannot solve the 2nd one (I'm ok with 3 and 4), how to do it? and is the brightness of a light bulb determined by its power? If two two light bulbs of different power are connected in series, will they have different brightness?

回答 (1)

天同

2011-09-20 7:36 am

✔ 最佳答案

Question No.1 - 5
The most simple way is to gegin from the starting point, and trace how many ways you could go through the resistors to the end point. You need to spend some effort in re-drawing the circuits.

6. The deflection indicates a current is passing through the galvanometer.
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1. Electric field tends to pass through metals because of its low resistance. Also, there should not be ay field inside the sphere because of constant potential on the sphere.

2. 'Positive induced charges are distributed uniformly on the outer surface of the shell.' Why is it correct?

The effect of induced -ve charge on the inner surface of the shell and the +ve charge at the cavity cancel out each other, leaving the +ve charge residing on the outer surface of the shell. Because of symmetry (the curvatures on any pointon the shell are the same) , charges distribute evenly on the shell surface.

'The electric field at any point within the cavity is zero' Why incorrect?
There is a +ve charge in the cavity. Field lines originate from the +ve charge and end up at the -ve induced charges on the inner shell surface.

'The electric potential at the outer surface of the shell is higher than that at its inner surface.' Why incorrect?
The shell is a conductor. Every point on it is of the same potential.

3. Where is your diagram ?
It seems that if the path is not along an equipotential surface, there is electric force acting on it, which would give an acceleration.

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