Question

Nine wires each of resistance $r=5\Omega$ are connected to make a prism as shown in figure. The equivalent resistance of the arrangement across $\text{AB}($in $\Omega )$ will be :

• A. $6\Omega$
• B. $5\Omega$
• C. $1\Omega$
• D. $3\Omega$

Answer 1

The correct option is D $3\Omega$

We can infer that there is a line of symmetry through which a plane passes, which divides the circuit into two mirror image parts.


As per the current distribution shown in above figure, there will be no current passing through wires $\text{DE}$ & $\text{CF}$ because whatever current passing through $\text{AE}$, due to symmetry same amount of current will be flow from $\text{FB}$ and similarly current won't flow through $\text{CF}$.

Hence, their resistance can be neglected. Now redrawing the circuit,


The equivalent resistance for the three resistance $r$ in series & again in parallel with resistance $r$ connected between $\text{A}$ and $\text{B}$ is :

$r^{\prime }=\frac{\left(3r\right)\left(r\right)}{3r+r}=\frac{3r}{4}$


Now the effective resistance between $\text{A}$ and $\text{B}$ is given by

$R_{eq}=\frac{\left(\frac{3r}{4}\right)\left(3r\right)}{\frac{3r}{4}+3r}$

$R_{eq}=\left(\frac{9r^2}{4}\right)/(\frac{15r}{4})=\frac{3r}{5}$

Given that $r=5\Omega$

$\therefore R_{eq}=\frac{3r}{5}=\frac{3\times 5}{5}=3\Omega$

Why this question? Tip: The given prism is a regular prism because the edge length are equal and resistance of each wire is $5\Omega$. Thus, we can observe a line of symmetry (through which a plane passes) and it will lead of equal current in identical branches.