Consider the circuits shown in fing Both the circuits are taking same current from the battery, but the current through R in the second circuit is 1//10 th of the current thourgh R in the first cirucit , if R is `11Omega` then find the value of `R_1 and R_2`

Explanation

From the circuit given in question , current through `R_(2)` is
`i-(i)/(10) = (9i)/(10)`
Potential difference across `R_(2)` is equal to potential difference
across R. Therefore,
`R_(2)xx(9)/(10)i= Rxx(i)/(10), i.e., R_2 = (R )/(9) = (11)/(9) Omega`
`:. R_(eq) = (R_(2) xxR)/(R_(2)+R) = ((11)/(9) xx(11)/(1))/((11)/(9) +(11)/(9)) = (11)/(10)Omega`
Toal circuit resistance is
`(11)/(9) +R_(1) = R =11`
or `R_(1) = 11 - (11)/(10) = 9.9 Omega`