The circuit in figure, where there are three resistors `R_(1),R_(2)` and `R_(3)` . If the voltage between A and C is V, the current / is given by

Options

A. `(V(R_(1) +R_(2)))/(R_(1)R_(2) + R_(1)R_(3) + R_(2)R_(3))`
B. `(V(R_(1)+R_(3)))/(R_(1)R_(2)+R_(1)R_(3)+R_(2)R_(3))`
C. `(V(R_(3)xxR_(2)))/(R_(1)R_(2)+R_(1)R_(3)+R_(2)R_(3))`
D. `(V(R_(2)+R_(3)))/(R_(1)R_(2)+R_(1)R_(3)+R_(2)R_(3))`

Answer

D

Explanation

We can replace `R_(2)` and `R_(3)` by an equivalent `R_(eq)^(23)` between point B and C with

The circuit now has `R_(1)` and `R_(eq)^(23)` in series and hence their combination can be replaced by an equivalent resistance `R_(eq)^(123)` with
`R_(eq)^(123)=R_(eq)^(23)+R_(1)=R_(1)+[(R_(2)R_(3))/(R_(2)+R_(3))]`
If the voltage between A and C is V, the current l is given by
`l = (V)/(R_(e )^(123))=(V)/(R_(1)+[R_(2)R_(3)//(R_(2)+R_(3))])`
`= (V(R_(2)+R_(3)))/(R_(1)R_(2)+R_(1)R_(3)+R_(2)R_(3))`