Question

Find the equivalent resistance of the circuits shown in figure between the points a and b. Each resistor has a resistance r.

Answer 1

a) $R_{eff}=\frac{\left(2r/2\right)\times r}{\left(2r/2\right)+r}$

$=\frac{r^2}{2r}=\frac{r}{2}$

b) At $0$ current coming to the junction is current foing from $BO=$ Current going along $OE$.

Current on $CO=$ Current on $OD$

Thus it can be assumed that current coming in $OC$ goes in $OB$.

Thus the figure becomes

$[r+\left(\frac{2r.r}{3r}\right)+r]=2r+\frac{2r}{3}=\frac{8r}{3}$

$R_{eff}=\frac{\left(8r/6\right)\times 2r}{\left(8r/6\right)+2r}=\frac{8r^2/3}{20r/6}=\frac{8r^2}{3}\times \frac{6}{20}=\frac{8r}{10}=4r$.