Question

Eight identical resistance each $15\Omega$ are connected along the edge of a pyramid having square base as shown. The equivalent resistance between $AandD$ is :

• A. $7\Omega$
• B. $8\Omega$
• C. $\frac{15}{2}\Omega$
• D. $\frac{15}{4}\Omega$

Answer 1

The correct option is B $8\Omega$

Given: Eight identical resistance each $15\Omega$ are connected along the edge of a pyramid having square base.

To find the equivalent resistance between Aand D

Solution:

Let current I be drawn by the battery, then $R=\frac{E}{I}$

Suppose the current $I$ enters at point A, a part $I_1$ flow along AO, part $I_2$ flow along AB and current $(I-I_1-I_2)$flow along path AD. (as shown in above figure).

Also when the current $I_2$ enter at point B, its part $I_3$ flows along BO and he remaining part $(I_2-I_3)$ will flow along DC. Again by symmetry, the current along the path OC will be $I_3$ and current along path CD will be $I_2$.

By symmetry the current $I$ flows out from D will be composed of part $I_1$ from OD, part $I_2$ from CD and part $(I-I_1-I_2)$ from path AD.

Applying Kirchhoff's second law to closed loop $ADOA$,

$-(I-I_1-I_2)\left(15\right)+I_1\left(15\right)+I_1\left(15\right)=0⟹3I_1+I_2=I.........\left(i\right)$

For closed loop ABOA,

$-I_2\left(15\right)-I_3\left(15\right)+I_1\left(15\right)=0⟹I_1-I_2-I_3=\mathrm{0....}\left(ii\right)$

For closed loop CBOC,

$(I_1-I_3)\left(15\right)-I_3\left(15\right)-I_3\left(15\right)=0⟹I_2-3I_3=\mathrm{0....}\left(iii\right)⟹I_3=\frac{I_2}{3}....\left(iv\right)$

Substituting the value of $I_3$ in equation (ii) we get,

$I_1-I_2-\frac{I_2}{3}=0⟹I_1=\frac{4}{3}{I}_2....\left(v\right)$

Substituting the value of $I_1$ in equation (i) we get,

$(3\times \frac{4}{3}{I}_2)+I_2=I⟹5I_2=I⟹I_2=\frac{I}{5}....\left(vi\right)$

From equation (v) and (vi), we get

$I_1=\frac{4}{3}{I}_2=\frac{4}{3}\times \frac{I}{5}=\frac{4I}{15}$

In closed loop ADYXA of the circuit,

$(I-I_1-I_2)\left(15\right)-E=0⟹(I-I_1-I_2)\left(15\right)=E⟹(I-\frac{4I}{15}-\frac{I}{5})\left(15\right)=E⟹\frac{8}{15}I\left(15\right)=E⟹E=8I$

So, $R_{eq}=\frac{E}{I}=\frac{8I}{I}=8\Omega$