I'll line up some tough ones now. These are definitely NOT for the weak-of-heart.
'n' points in 3D space are connected to each other by capacitors. (Every point is connected to every other point.)
Two points are selected at random. What's the capacitance b/w these two points?
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ReplyDeletenC/2 ?
ReplyDeletenC/2
ReplyDeleteby checking values at n=2,3,4
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ReplyDeleteall points except the terminals will be at same potential, so, effectively, there is one capacitor b/w the terminals, and (n-2) capacitors parallel with that
ReplyDeleteC + (n-2)C/2=nC/2
imagine this:
ReplyDeletei draw a hexagon and connect all its vertices with capacitors. (and i don't state that 'n' points thing.)
not so simple then???
well, i learned it the hard way.
please elaborate some more @ sambhav and bhaiya
ReplyDeletein that case, i would have formed wheat stone bridges(nC/2 won't click, and neither will it seem to be an obvious answer)
ReplyDeleteso, there would be two wheat stone bridges of effective capacitance C, and one capacitance between the terminals itself, hence answer 3C
is there a better way?
how should i know?? i couldn't solve it.
ReplyDeleteso, is nC/2 correct?
ReplyDeleteyup..
ReplyDelete@sambhav: How can u ignore the fact that the n-2 capacitors which u've considered in parallel will also be connected amongst themselves????
ReplyDeletesambhav please answer prateek, i have same doubt
ReplyDeletewell , i have nw understood the thing......as all the points except the two points we have considered are symmetrically placed....so all of them wud hv the same potential.......so we can remove all the capacitances in b/w those points.....
ReplyDelete@ delhiite
ReplyDelete"How can u ignore the fact that the n-2 capacitors which u've considered in parallel will also be connected amongst themselves????"
we selected any two points at random,now apart from these two points ,all are other points are symetrical(if u dont agree on this,tell me why 1 point will be more spl than the other)
now any capacitor joined between to equipotential pts is useless,so if we had selected A and B as our pts and other pts are C1,C2,C3.....C(n-2)
then effective connections left in the circuit are
A-C1-B , A-C2-B , A-C3-B .........A-C(n-2)-B
=>n-2 connections in parallel