A wire is bent in the form of circle of radius 2m. Resistance per unit length of wire is `(1)/(pi) (Omega)/(m)`. Battery of 6 V is connected between A and B `angle AOB = 90^(@) ` . Find the current through the battery.

Options

A. `8 A`
B. `3 A`
C. `4 A`
D. `9 A`

Answer

A

Explanation

`8 A`
Length of wire l ` = 2 pi r = 2 pi xx 2 = 4 pi ` meter Now, total resistance R = `rho l = 4pi xx (1)/(pi ) = 4 Omega`
Resistance of smaller chord `R_(1) = (4 xx 90)/(360) = 1 Omega`
Resistance of bigger chord `R_(2) = 4 - 1 = 3 Omega`
`therefore ` Equivalent resistance between A and B
` R =(R_(1)R_(2))/(R_(1) + R_(2)) `
` therefore R = (1 xx 3 )/(1 + 3) = (3)/(4) Omega` ,
`therefore` Current flowing through circuit `I = (V)/(R) = (6)/((3)/(4))`
`therefore I = (24)/(3) `
`therefore I = 8 A `.