question_answer

Five resistance of resistance RQ are these, 3 are connectd in parallel and are joined to them in series. Find resultant resistance:

A)  \[\left( \frac{3}{7} \right)R\,\Omega \]             

B)  \[\left( \frac{7}{3} \right)R\,\Omega \]

C)  \[\left( \frac{9}{8} \right)R\,\Omega \]      

D)  \[\left( \frac{8}{9} \right)R\,\Omega \]

Correct Answer: B

Solution :

From figure first three resistance are connected in parallel. \[\Rightarrow \] \[\frac{1}{{{R}_{P}}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R}\] \[\Rightarrow \] \[{{R}_{P}}=\frac{R}{3}\Omega \] Now, the reduced diagram will be. All the resistance are connected in series. \[\therefore \] \[{{R}_{eff}}=\frac{R}{3}+R+R=\frac{R}{3}+2\Omega \] \[\Rightarrow \] \[{{R}_{eff}}=\frac{7R}{3}\Omega .\]