JEE Main 26-Aug-2021 Shift 2 Solved Paper

The given value of resistances are 2Ω,4Ω,6Ω and 8Ω.
The required value of combination is

46

3

Ω.
In order to achieve the above mentioned values of resistance from given resistances, we will connect 2Ω and 4Ω resistance in parallel, then join 6 Ω and 8 Ω resistance in series with the combination.
The circuit diagram for connection is shown below. ∴Req=(2|4)+6+8=

2×4

2+4

+14=

46

3

Ω
Thus, resistance of 2Ω and 4Ω are in parallel with 6Ω and 8Ω in series combination.