Solve this:
Q. The equivalent resistance of the network shown in the figure between the points A and B is
(A) 4r
(B) 4r/3
(C) 3r
(D) 3r/4

Dear Student ,
Here in this case the three resistors of resistance r are in parallel with each other and then they are in series with resistor r beside the point a .
So here in this case the equivalent resistance of the circuit is ,
$\frac{1}{{\displaystyle \frac{1}{r}+\frac{{\displaystyle 1}}{{\displaystyle r}}+\frac{{\displaystyle 1}}{{\displaystyle r}}}}+r=\frac{r}{3}+r=\frac{4r}{3}$
Hope this helps you .
Regards