Question

Arrange 8, 12 & 16 12 Resistance such that Req=8.89 ​

Answer 1

By arranging, 8Ω and 12Ω in a series connection and then the connection of the resultant in parallel with 16Ω.

When arranging the 8Ω and 12Ω in a series,

8 Ω + 12Ω= 20 Ω

Now, taking this 20Ω and arranging it in parallel combination with the 16Ω,

we get,

$\frac{1}{r}$ = $\frac{1}{20}$ + $\frac{1}{16}$

$\frac{1}{r}$ = $\frac{9}{80}$

r = $\frac{80}{9}$

∴ r = 8.89 Ω