In the electrical circuit shown in the figure, the current i through the side AB is .....

Options

A. `(6)/(25)`A
B. `(10)/(33)`A
C. `(1)/(5)` A
D. `(10)/(63)`A

Answer

A

Explanation

`(6)/(25)A`

Equivalent resistance of parallel connection of `20 Omega and 30 Omega`
`R_(1) = (20 xx 30)/(20 + 30)`
`= (600)/(50) = 12 Omega`
Now, equivalent resistance of series connection of `10 Omega, R_(1) and 3 Omega`
R = 10 + `R_(1) + 3 `
`= 10 + 12 + 3 `
`therefore R = 25 Omega`

`therefore I = (10)/(25 ) = 0.4 A `
At A, I = 0.4 is distributed in two parts.
Suppose, current flowing through `20 Omega` is `I_(1)` and `30 Omega ` is `I_(2)`
`therefore ` Current flowing through `20 Omega`
`i = I_(1) = I xx (R_(2))/(R_(1) + R_(2)) `
`i = 0.4 xx (30)/(20 + 30)`
`i = (12)/(50) = (6)/(25) `A