Six resistors each of resistance R and two resistors each of resistance r are connected in the network shown below. The equivalent resistance between A and B is

Options

A. 3R+r
B. 6R+2r
C. 2R/3
D. `R/3+r/3`

Answer

C

Explanation

`R_1` and `R_2` are in series , similarly `R_3` and `R_4` are in series, `R_5` and `R_6` are also in series . All having equal value of resistance i.e., R

This complete setup is in parallel
So, `1/(R_"(effective)")=1/"2R"+1/"2R"+1/"2R"`
`1/(R_"(effective)")=(1+1+1)/"2R"=3 /"2R"`
`R_"effective"=(2R)/3`
Here, no current will flow in 2r(r+r), because this will behave as null point circuit .