Question

the following combination has an effective resistance of 8ohm calculate the value of 'R'

Answer 1

Given that the equivalence resistance of the circuit is 8Ω.

Here,

$\sf R_1$ and $\sf R_2$ are in parallel.

Thus,

$\sf \dfrac{1}{ R_1} + \dfrac{1}{ R_2} = \dfrac{1}{ R_p} \\ \\ \longrightarrow \sf \: \dfrac{1}{ R_p} = \dfrac{1}{10} + \dfrac{1}{8} \\ \\ \longrightarrow \sf \: \dfrac{1}{ R_p} = \dfrac{9}{40} \\ \\ \longrightarrow \sf \: R_p = \dfrac{9}{40}$

Here,

$\sf R_p$ and R are connected in series.

Thus,

$\longrightarrow \sf R_p + R = 8 \\ \\ \longrightarrow \sf \ R = 8 - \dfrac{40}{9} \\ \\ \longrightarrow \sf \ R = 8.00 -4.44 \\ \\ \longrightarrow \boxed{ \boxed{\sf \ R = 3.56 \Omega}}$