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 )/(2pir)`
Length of sections ADB and ACB are `rtheta and r theta and (2pir-theta)`
Resistance of section ADD, `R_(1)=rhortheta=(R)/(2pir)rtheta=(Rtheta)/(2pi)`
and resistance of section, ACB,
`R_(2)=rhor(2pi-theta)=(R)/(2pir)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))=((Rtheta)/(2pi)xx(R(2pi-theta))/(2pi))/((Rtheta)/(2pi)+R(2pi-theta)/(2pi))=(Rtheta(2pi-theta))/(4pi^(2))`
`"Putting" theta=45^(@)=(pi)/(4) "rad" and R=15Omega `
`R_("eq")=((15xx(pi))/(4)xx(2pi-(pi)/(4)))/(4pi^(2))=((15)/(4)((7pi)/(4)))/(4pi^(2))=1.64Omega`