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A copper wire is in the form of a cylinder and has a resistance R. it is stretched till its thickness reduces by half of its initial size. Find its new resistance in terms of R.
its resistance will be increased.because R is always inversely proportional to lengthA copper wire is in the form of a cylinder and has a resistance R. it is stretched till its thickness reduces by half of its initial size. Find its new resistance in terms of R.
Thanks, could you please mention by how much the resistance will rise ?its resistance will be increased.because R is always inversely proportional to length
I am really sorry to confuse you but this question appeared for my board exams recently . Could you please help me out .This should probably go into the homework section, and you will need to show your work so far.
R is http://en.wikipedia.org/wiki/Electrical_resistance#DC_resistance" inversely proportional to length.its resistance will be increased.because R is always inversely proportional to length
it resistance will be increased by two times.Thanks, could you please mention by how much the resistance will rise ?
i mean 9its diameterR is http://en.wikipedia.org/wiki/Electrical_resistance#DC_resistance" inversely proportional to length.
Are you sure? Aren't radius and diameter directly proportional, so that if you decrease the diameter by a half, you also decrease the radius by a half?[tex] R = \rho \frac{h}{A}[/tex]
We must first notice that both area and length change so that the volume remains constant. We assume that the stretching results in a prefect cylinder that we began with. Therefore:
[tex] V = Ah = \pi r^2 h[/tex] and [tex] V[/tex] is a constant.
Thickness refers to the diameter of the cylinder, so to decrease its thickness by half, you must decrease its diameter by half (or its radius by a fourth). So the new area is:
[tex] A' = \pi (\frac{r}{4})^2 = \pi \frac{r}{16}[/tex]
Therefore, to destroy no matter and to keep volume constant, we must multiply [tex]h'[/tex] by a constant to keep [tex]V'[/tex] equal to [tex]V[/tex]:
[tex] V = V' = \pi r^2 h = (\pi \frac{r^2}{16})(Kh)[/tex]
Seen above, K must equal sixteen to cancel out that fractional sixteen, thus bringing both left and right equations equal and thus keeping volume constant.
To sum:
[tex] h' = 16h[/tex]
[tex] A' = \pi \frac{r^2}{16} = (\pi r^2) \frac{1}{16} = \frac{A}{16}[/tex]
We now plug in our new height and our new area into the equation for resistance:
[tex] R' = \rho \frac{16^2h}{A}= 16^2(\rho \frac{h}{A}) = 16^2 R[/tex]
you're rightAre you sure? Aren't radius and diameter directly proportional, so that if you decrease the diameter by a half, you also decrease the radius by a half?