Question

Please answer the (ii) part of Q29.

Answer 1

Answer:

just a sec solving

Explanation:

so the resistances are 4 and 12

Answer 2

Answer:

by the given,

when connected in series,

$r _{1} + r _{2} = 16 \: \: \: \: \: \: .......1$

when connected in parallel,

$\frac{1 }{r _{1} } + \frac{1}{r _{2} } = \frac{1}{3} \\ \frac{r _{1} + r _{2} }{r _{1}r _{2} } = \frac{1}{3} \\ \frac{16}{r _{1} r _{2}} = \frac{1}{3 } \\ 48 = r _{1} r _{2} \\ \\ ({r _{1} - r _{2}) }^{2} = ({r _{1} + r _{2} })^{2} - 4r _{1} r _{2} \\ ( {r _{1} - r _{2} )}^{2} = {16}^{2} - 4(48) \\ r _{1} - r _{2} = \sqrt{256 - 192} \\ {r _{1} - r _{2} } = 8 \: \: \: \: \: \: \: \: \: \: \: ......2$

From 1 and 2 we have ,

R1=12 ohm and R2=4 ohm

hope it helps you...

please mark it as brainliest....