Answer: [tex]200\Omega[/tex]
Explanation:
Given
The battery used is 6V in magnitude
The current flow is 90 mA
If the resistance of A is [tex]100\Omega[/tex]
Net resistance of the circuit is
[tex]R_{net}=\dfrac{V}{I}\\\\\Rightarrow R_{net}=\dfrac{6}{90\times 10^{-3}}\\\\\Rightarrow R_{net}=\dfrac{200}{3}\Omega[/tex]
If A and B are in parallel then their net resistance must be [tex]\frac{200}{3}[/tex]
[tex]\dfrac{1}{\frac{200}{3}}=\dfrac{1}{100}+\dfrac{1}{R_b}\\\\\Rightarrow \dfrac{1}{R_b}=\dfrac{3}{200}-\dfrac{1}{100}\\\\\Rightarrow \dfrac{1}{R_b}=\dfrac{3-2}{200}\\\\\Rightarrow R_b=200\Omega[/tex]
