In above circuit if current through 10 `Omega` resistor is 2.5 A, value of R is .... .

Options

A. `40 Omega`
B. `10 Omega`
C. `8 Omega`
D. `50 Omega`

Answer

C

Explanation

`8 Omega`
Suppose, `R_(1) = 10 Omega, R_(2) = 40 Omega`

For parallel connection of 10`Omega and 40 Omega`,
`I_(1) = I xx (R_(2))/(R_(1) + R_(2))`
` therefore I = I_(1) xx (R_(1) + R_(2))/(R_(2))`
` = 2.5 xx (10 + 40)/(40)`
`therefore I = 3.125` A
Equivalent resistance of parallel connection of 10 `Omega and 40 Omega`,
`R. = (10 xx 40)/(10 + 40)`
` = (400)/(50) = 8 Omega`
and current in circuit ` I = (E)/(R. + R)`
3.125 = `(50)/(8 + R)`
`therefore 8 + R = (50)/(3.15) `
`therefore 8 + R = 16 " " therefore R = 8 Omega`