Question

Find the equivalent resistance of the circuits shown in the figure (32-E17) between the points a and b. Each resistor has resistance r.

Answer 1

(a) The simplified circuit can be drawn as shown below.

Here cdeo forms a balanced Wheatstone bridge; therefore, branch od will become become ineffective.
The simplified circuit will then be as shown below.

The equivalent resistance between points c and e,
$R_{cd}=\frac{2r\times 2r}{4r}=r$
The equivalent resistance between a and b,
$R_{ab}=\frac{r\times r}{2r}=\frac{r}{2}$
(b)


Let R_eff be the effective resistance of the circuit.Then, from the symmetry of the circuit, we can assume that the current moving along CO enters OB and the current moving along EO enters OD.
Current on CO = current on OB
Current on EO = current on OD
So, the circuit can be simplified as shown below.


From the simplified circuit diagram, effective resistance of the upper half of the circuit will be
$$\begin{gathered} \left[\left[r+\left(\frac{2r^2}{3r}\right)+r\right]=2r+\left(\frac{2r}{3}\right)=\frac{8}{3}r\right] \\ \Rightarrow {\mathrm{R}}_{\mathrm{eff}}=\frac{{\displaystyle \frac{8r}{6}}\times 2r}{{\displaystyle \frac{8r}{6}}+2r} \\ =8r^2\times \frac{2}{20r}=\frac{8r}{10}=\frac{4r}{5} \end{gathered}$$