Eight identical resistors r, each are connected along the edges of a pyramid having square base ABCD as shown in figure below. Calculate equivalent resistance between A and B. Solve the problem (i) without using Kirchhoff’s laws (ii) by using Kirchhoff’s laws.

Eight identical resistors 'r' are connected along the edges of a pyramid and we have to calculate the equivalent resistance.



(i) Without using Kirchhoff’s laws:  

Consider a battery connected between A and B. The circuit now has a plane of symmetry. This plane of symmetry passes through the mid-points of AB and CD and the vertex O.
So, the currents are same in (i) AO and OB (ii) DO and OC.
Now, OA and OB can be treated as a series combination which gives resistance 2r.
Also, DO and OC are in series combination giving resistance 2r.
It is in parallel with DC.
This gives a resistance of $\frac{2\mathrm{r}\times \mathrm{r}}{2\mathrm{r}+\mathrm{r}}$ = $\frac{2{\mathrm{r}}^2}{3\mathrm{r}} = \frac{2\mathrm{r}}{3}.$

This is in series with resistances AD and CB.

Hence, their combined resistance is $\frac{2\mathrm{r}}{3}+2\mathrm{r}$= $\frac{8\mathrm{r}}{3}.$ 

Now $\frac{8\mathrm{r}}{3},$ resistor AB (r) and combination of AO and BO (i.e., 2r) are in parallel.

If R is the equivalent resistance, then
                    $$\begin{gathered} \frac{1}{\mathrm{R}} = \frac{3}{8\mathrm{r}}+\frac{1}{\mathrm{r}}+\frac{1}{2\mathrm{r}} \\ = \frac{3+8+4}{8\mathrm{r}} \\ = \frac{15}{8\mathrm{r}} \end{gathered}$$

$\Rightarrow$                  $\mathrm{R} = \frac{8\mathrm{r}}{15}$

(ii) Using Kirchhoff’s laws: 

Using Kirchhoff’s second law in loop DOCD, we get

– I₃r – I₁r + (I₂ – I₃)r = 0
$\Rightarrow$ – 3I₃r + I₂r = 0
$\Rightarrow$               ${\mathrm{I}}_3 = \frac{{\mathrm{I}}_2}{3}$                            ...(i) 

Again, using Kirchhoff’s loop law to loop AOBA, we get

– I₁r – I₂r + (I – I₁ – I₂) r = 0
$\Rightarrow$ 3 I₁ + I₂ = I                                  ...(ii) 

Considering loop ADCBA, we get

$-{\mathrm{I}}_2\mathrm{r}-({\mathrm{I}}_2-{\mathrm{I}}_3)\mathrm{r}-{\mathrm{I}}_2\mathrm{r}+(\mathrm{I}-{\mathrm{I}}_1-{\mathrm{I}}_2)\mathrm{r} = 0$

$\Rightarrow$$\mathrm{Ir}-{\mathrm{I}}_1\mathrm{r}-4{\mathrm{I}}_2\mathrm{r}+{\mathrm{I}}_3\mathrm{r} = 0$

$$\begin{gathered} \Rightarrow \mathrm{I} = {\mathrm{I}}_1+4{\mathrm{I}}_2-{\mathrm{I}}_3 \\ \Rightarrow \mathrm{I} = {\mathrm{I}}_1+4{\mathrm{I}}_2-\frac{{\mathrm{I}}_2}{3} \end{gathered}$$ 

Using equation (i), we get

  $\mathrm{I} = {\mathrm{I}}_1+\frac{11}{3}{\mathrm{I}}_2$
Using equation (ii), we get 

 $3{\mathrm{I}}_1+{\mathrm{I}}_2 = {\mathrm{I}}_1+\frac{11}{3} {\mathrm{I}}_2 \mathrm{or} {\mathrm{I}}_2 = \frac{3}{4}{\mathrm{I}}_1$

From equation (ii),

$\mathrm{I} = 3 {\mathrm{I}}_1+\frac{3}{4}{\mathrm{I}}_1 = \frac{15}{4}{\mathrm{I}}_1$

Considering circuit ABEA,

$\mathrm{E}-(\mathrm{I}-{\mathrm{I}}_1-{\mathrm{I}}_2)\mathrm{r} = 0$

$$\begin{gathered} \Rightarrow \mathrm{E} = (\mathrm{I}-{\mathrm{I}}_1-{\mathrm{I}}_2) \mathrm{r} \\ = \left(\frac{15}{4}{\mathrm{I}}_1-{\mathrm{I}}_1-\frac{3}{4}{\mathrm{I}}_1\right)\mathrm{r} \end{gathered}$$ 

$\therefore$ E = 2 l₁r                               ...(iii) 

If R is the total resistance, then E = I R

i.e.,              $\mathrm{E} = \frac{15}{4}{\mathrm{I}}_1\mathrm{R}$
                $2{\mathrm{I}}_1\mathrm{r} = \frac{15}{4}{\mathrm{I}}_1\mathrm{R} (\mathrm{from} (\mathrm{iii}\left)\right)$
$\therefore$          $$\begin{gathered} \frac{15}{4}\mathrm{R} = 2\mathrm{r} \\ \mathrm{R} = \frac{8\mathrm{r}}{15}. \end{gathered}$$  
which is the required equivalent resistance using Kirchoff's law.