Question

Three resistors of 2 ohm 3
ohm and 6ohm are give what is the least resistance that you can get using all of them​

Answer 1

GiveN :

Three resistors are given with magnitude as $\sf{2 \Omega \: , \: 3 \Omega \: and \: 6 \Omega}$ respectively.

To FinD :

Least value of the three resistances.

SolutioN :

For the maximum value of the resistances we've to connect all of them in the $\tt{\green{Series \: combination}}$ , where as for calculating the minimum value of the resistance we've to connect then in $\tt{\green{Parallel \: combination}}$ .

Let,

$\rm{R_1 = 2 \Omega}$

$\rm{R_2 = 3 \Omega}$

$\rm{R_3 = 6 \Omega}$

Add them in parallel for getting minimum value :

$\implies \rm{\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}} \\ \\ \\ \implies \rm{\dfrac{1}{R_{eq}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}} \\ \\ \\ \implies \rm{\dfrac{1}{R_{eq}} = \dfrac{3 + 2 + 1}{6}} \\ \\ \\ \implies \rm{\dfrac{1}{R_{eq}} = \dfrac{6}{6}} \\ \\ \\ \implies \rm{\dfrac{1}{R_{eq}} = 1} \\ \\ \\ \large \implies {\boxed{\rm{R_{eq} = 1 \Omega}}}$

Answer 2

R¹ = 2Ω

R² = 3Ω

R³ = 6Ω

Add them in parallel for getting minimum value :

$\begin{gathered}\hookrightarrow \sf{\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}} \\ \\ \\ \hookrightarrow \sf{\dfrac{1}{R_{eq}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}} \\ \\ \\ \hookrightarrow \sf{\dfrac{1}{R_{eq}} = \dfrac{3 + 2 + 1}{6}} \\ \\ \\ \hookrightarrow \sf{\dfrac{1}{R_{eq}} = \dfrac{6}{6}} \\ \\ \\ \hookrightarrow\sf{\dfrac{1}{R_{eq}} = 1} \\ \\ \\: \large \implies {\boxed{\underline{\rm{\red{R_{eq} = 1 \Omega}}}}}\end{gathered}$