Question

What is the (a) highest (b) lowest resistance that can be secured by combining four coils of resistance $4\Omega ,8\Omega ,12\Omega$, and $24\Omega$

Answer 1

$\left(a\right)$Highest will be by series $=4+12+8+24=48ohms$
$\left(b\right)$Lowest will be by parallel $=1/R=1/4+1/8+1/24+1/12=12/24=1/2$
so $R=2$