Question

A resistance of 6Ω is connected in parallel with a 3Ω resistance. Both resistance are then connected
in series with 8Ω resistance. How much is the total resistance of the circuit?

Answer 1

Gives

$R_1=6\Omega\\R_2=3\Omega\\R_3=8\Omega\\$

$R_{net}=?$

$R'_{net}=\frac{R_1R_2}{R_1+R_2}=\frac{6\times3}{6+3}=2\Omega$

$R_{net}=R'_{net}+R_3=2+8=10\Omega$