Consider the circuit shown in the figure. Both the circuits are taking same current from battery but current through `R` in the second circuit is `(1)/(10) th` of current through `R` in the first circuit. If `R` is `11 Omega`, the value of `R_(1)`

Options

A. `9.9 Omega`
B. `11 Omega`
C. `8.8 Omega`
D. `7.7 Omega`

Answer

A

Explanation

(a) In figure (b) through `R_(2) = i- (i)/(10) = (9 i)/(10)`
Potential difference across `R_(2) =` Potential difference
across `R implies R_(2) xx (9)/(10) I = R xx (1)/(10)`, i.e.,
`R_(2) = (R )/(9) = (11)/(9) Omega`
`R_(eq) = (R_(2) xx R)/((R_(2) + R)) = ((11)/(9) xx (11)/(1))/((11)/(9) + (11)/(1)) = (11)/(10) Omega`
Total circuit resistance `= (11)/(10) + R_(1) = R = 11`
`implies R_(1) = 9.9 Omega`