IIT JEE , AIEEE Forum - Physics , Electricity

caprio.nups

A and B are two points on a uniform ring of resistance R. The

ACB =

, where C is the centre of the ring. The equilvalent resistance b/w A and B is

a. ( R / 4

²) (2

-

)

b. R [ 1 - (

/ 2

) ]

c. ( R

) / 2

d. R [ (2

-

) / 4

]

rohitju

it is equivalent to 2 resistances in paralell. one resistance is the minor arc & the other is the major arc.then apply laws of resistances for parallel config. let radius of circle is r.then length of minor arc is r.(theta)

length of major arc is r.(360-- theta)

resistance per unit length is R/2

r

vikalp

hey..........

the resistance per unit length will be equal to R/2

r .....(let the radius of ring be r)

so if you draw a figure then the major arc AB and the minor arc AB will be in parellel and the circuit will be equivalent to two resistances in parellel ....

let these two bE R1 and R2 respectively..............

R₁= ( R/2r )*(2-)*(r)

and R₂ = (R/2r)*()*(r)

{ resistance per unit length multiplied by lengths }

so now equivalent reistance will be :

R₀ =R₁R₂/R₁+R₂

and hence R₀ = ( R / 4

²) (2

-

)

...................................

keep it cooooolll...............

atul_sinha89

consider the minor and major arc as two parallel resistors, lert radius of the ring be r ,then

resistance per unit length (RP)= R / 2(pi)r
resistance of minor arc = [RP]*r(theta)

resistance of major arc = [RP]* r[2(pi)-(theta)]

calculate the eqivalent resistance for parallel resistors of above resistance..

ans=(a)

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