A and B are two points on a uniform ring of resistance `15Omega`. The `ltAOB=45^(@)` The equivalent resistance between A and B is

Options

A. `1.64Omega`
B. `2.84Omega`
C. `4.57Omega`
D. `2.64Omega`

Answer

A

Explanation

Resistance per unit length of ring, `rho=(R )/(2pi r)`
Length of sections ADB and ACB are `r theta and r(2pi-theta)`
`:.` Resistance of section ADB: `R_(1)=rho r theta`
`=(R )/(2pi r) r theta =(R theta)/(2pi)`
and resistance of section ACB, `R_(2)= rho r(2pi-theta)`
`=(R )/(2pi r) r(2pi-theta) =(R (2pi-theta))/(2pi)`
Now, `R_(1) and R_(2)` are connected in parallel between A and B then
`R_(eq)=(R_(1)R_(2))/(R_(1)+R_(2)) = ((R theta)/(2pi) xx (R(2pi -theta))/(2pi))/((R theta)/(2pi)+(R(2pi-theta))/(2pi))`
`((R theta xx R (2pi -theta))/((2pi)^(2)))/((R theta +R(2pi-theta))/(2pi))=(R theta (2pi-theta))/(4pi^(2))`
Putting `theta =45^(@) =(pi)/(4) ` rad and `R=15Omega`
`R_(eq)=(15 xx (pi)/(4) xx (2pi -(pi)/(4)))/(4pi^(2))=((15pi)/(4)((7pi)/(4)))/(4pi^(2))`
`=(105)/(64)Omega =1.64Omega`