IIT JEE , AIEEE Forum - Physics , Electricity

ambareesh13

solve hc verma q 10 pg 198refer for figure in the book (electric current in conductors)
i want the detailed soln
thanx in advance

ambareesh13

why isnt anyone replying????

obaid007in

hey i m realy sry 4 wht i did ! it wont be repeated agan!

ambareesh13

waterdemon

Obaid007

Hey dude why don't u log onto Orkut if u just want to do this.

If this is what u have come here for then plZ join orkut.

This place is for solving doubts for others and practicing

Your concepts. Get lost !!!!!!!!!!!!!!!

And ambareesh I suggest u do be away from such people.
Cheers !!!!!!!!!!!!!!!!!!! :-)

ambareesh13

yeh i thought first but if anyone goes after my family .i aint goin to leave them .....what would have you done if you were me??

waterdemon

I Have attached the diagram along with this for explanation

Ther is a ring with a small element .

resistivity =

Taking (b-a) << L

R =

L / A

R =

(

x /

r²⁾

R (Eq) =

R

R (Eq) = _{[0 ]}^{[L ]}

dx /

r²

Tan

= (b-a)/L

Above "r" = a + xTan

R (Eq) =

dx /

(a+ xTan

)²

R (Eq) = (

/

Tan

) [ 1 / (a + xTan

)] From 0 to L

R (Eq) =

/

Tan

[ 1 /( a + xTan

)] From 0To L

Therefore From above Tan

= (b-a)/L

R (Eq) =

/

Tan

[ (-1 / a) + (1/ a + b - a )

R (Eq) =

/

Tan

( b - a / ab )

R (Eq) = [

/

(b - a)] * [(b -a) / ab]

(b - a) gets cancelled and we get

R (Eq) =

L /

ab

Hence solved

plZ rate me for my efforts.

Cheers !!!!!!!!!!!!!!@@@@@@@@!!!!!!!!!!!!!!!!!!!!!!!!!

As Tan

= (b-a) / L and (b-a

Admin

dear ambareesh13 ,

if people are putting so much effort to reply, pleazzz put up the question as well, else the context is lost when everyone doesnt have the book while accessing the web.

thats why there was no expert reply till now.

cheers
~admin

obaid007in

hey i m extremly sry! i was jst joking! i assure u this will nt be repeated!! thnx a lot 4 correctng me!

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