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

Consider the circuit shown in the figure. Both the circuits are taking

1.	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)` A. `9.9 Omega`B. `11 Omega`C. `8.8 Omega`D. `7.7 Omega`
Answer» Correct Answer - A (a) In Figure `(b)` through `R_(2) = i-(i)/(10) =(9i)/(10)` Potentail difference across `R_(2) =` Potentail difference across `R implies R_(2) xx (9)/(10) i= R xx (i)/(10) implies 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`

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