A) \[\frac{4R}{3}\Omega \]
B) \[2R\Omega \]
C) \[R\Omega \]
D) \[\frac{2R}{3}\Omega \]
Correct Answer: D
Solution :
Resistance R bisecting the circuit can be neglected due to the symmetry of the circuit. Now there are four triangles Effective resistance of each triangle \[\frac{1}{R}=\frac{1}{R}+\frac{1}{2R}\] \[=\frac{2+1}{2R}=\frac{3}{2R}\] \[\therefore \] \[R=\frac{2}{3}R\] Now the given circuit reduced to Therefore, effective resistance between A and B, \[\frac{1}{{{R}_{AB}}}=\frac{1}{2R}+\frac{1}{2R}=\frac{1}{R}\] \[\Rightarrow \] \[{{R}_{AB}}=R=\frac{2R}{3}\]You need to login to perform this action.
You will be redirected in
3 sec