Question

Eight identical resistors, each of resistance $r$, are connected along edges of a pyramid having square base ABCD as shown in the figure. Calculate equivalent resistance
(a) between A and D
(b) between A and O

Answer 1

(a) When a battery is connected across terminals A and D, circuit becomes symmetric about a plane passing through mid points of edges BC and AD and vertex O. Therefore, current through different resistors will be as shown in figure
Since currents in AO an dOD is equal, therefore they may be assumed in series with each other. Similarly, BO and OC may be assumed to be in series.
Hence, the circuit may be assumed as shown in figure(ii). Therefore, equivalent resistance of each series combination (BO and OC, AO and OD) is equal to $2r$. Series combination of BO and OC is parallel with BC
Equivalent resistance of this combination
$=\frac{r\times 2r}{r+2r}=\frac{2r}{3}$
Now, this combination is in series, with resistors AB and CD. Therefore, equivalent resistance of this series combination
$=\frac{2r}{3}+2r=\frac{8}{3}r$
The above series combination, resistor AD and series combination of resistors AO and OD are parallel to each other. Hence, equivalent resistance $R$ of the circuit is given by
$\frac{1}{R}=\frac{3}{8r}+\frac{1}{r}+\frac{1}{2r}orR=\frac{8r}{15}$
(b) Now the battery is connected across the terminals A an dO so the circuit becmoes symmetric about a plane passing through edges AO and OC, and vertex O current through different resistors will be as shown in figure (iii)
Applying Kirchoff's law on mesh OCDO
$i_{2}r+\frac{i_2}{2}r-i_{1}r=0\therefore i_1=\frac{3}{2}{i}_2....\left(1\right)$
For mesh ODAD
$i_{1}r+(i_1+\frac{i_2}{2})r-(i-2i_1-i_2)r=0or4i_1+\frac{3}{2}{i}_2=i.....\left(2\right)$
From Eqs (1) and (2)
$i_1=\frac{1}{5}i;i_2=\frac{2}{15}i$
For mesh OBAVO
$i_{1}r+(i_1+\frac{i_2}{2})r-V=0$
Substituting values of $i_1$ and $i_2$
$\frac{7}{15}=V$
But equivalent resistance of the circuit $=\frac{V}{i}$
Equivalent resistance $=\frac{7r}{15}$