Question

please solve this????????​

Answer 1

Case i): Rac  

This forms the condition of a blanced wheetstone bridge( $\frac{2}{1} = \frac{4}{2}$ )

so, the potential difference between B and D is same so resistor of 3 Ohms can be neglected.

so equivalent resistance Rac = $\frac{1}{\frac{1}{(2+4)} \frac{1}{(1+2)} }$

                               ⇒     Rac = 2Ω

Case ii) Rbd

        ⇒   $Rbd = \frac{1}{\frac{1}{(1+2)} +\frac{1}{3}+\frac{1}{2+4} } = \frac{6}{5} = 1.2$Ω

Case iii) Rad

        ⇒  $Rad = \frac{1}{\frac{1}{1}+\frac{1}{2+\frac{1}{\frac{1}{3}+\frac{1}{(2+4)} } } } = \frac{4}{5} = 0.8$ Ω

Case iv) Rbc

       ⇒  $Rbc = \frac{1}{\frac{1}{4}+\frac{1}{2+\frac{1}{\frac{1}{3}+\frac{1}{(1+2)} } } } = \frac{14}{11}$ Ω

Hope this helps you !!