Question

If you are provided a set of resistances 2Ω, 4Ω, 6Ω and 8Ω. Connect these resistances so as to obtain an equivalent resistance of `46/3`Ω.

Options

• 4Ω and 6Ω are in parallel with 2Ω and 8Ω in series.  

• 6Ω and 8Ω are in parallel with 2Ω and 4Ω in series.

• 2Ω and 6Ω are in parallel with 4Ω and 8Ω in series. 

• 2Ω and 4Ω are in parallel with 6Ω and 8Ω in series.

Answer 1

2Ω and 4Ω are in parallel with 6Ω and 8Ω in series.

Explanation:

Net Resistance = `1/"R"_1+1/"R"_2+"R"_3+"R"_4`

= `1/2+1/4+6+8`

= `(2xx4)/(2+4)+14`

= `46/3`Ω