Question

What is the minimum resistance which can be made using five resistors each of $\frac{1}{5}\mathrm{\Omega }$?

• A. $\frac{1}{5}\mathrm{\Omega }$
• B. $\frac{1}{25}\mathrm{\Omega }$
• C. $\frac{1}{10}\mathrm{\Omega }$
• D. $25\mathrm{\Omega }$

Answer 1

The correct option is B

$\frac{1}{25}\mathrm{\Omega }$

Step 1: Given information

No. of resistors, $\mathrm{n}=5$.

Each resistance, $\mathrm{R}=\frac{1}{5}\mathrm{\Omega }$.

Step 2: Minimum resistance is obtained when resistors are connected parallel. Net resistance in parallel combination is given by

$\frac{1}{{\mathrm{R}}_{\mathrm{net}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}+\frac{1}{{\mathrm{R}}_4}+\frac{1}{{\mathrm{R}}_5}$

Substitute $\frac{1}{5}\mathrm{\Omega }$ for ${\mathrm{R}}_1$,$\frac{1}{5}\mathrm{\Omega }$ for ${\mathrm{R}}_2$, $\frac{1}{5}\mathrm{\Omega }$ for ${\mathrm{R}}_3$,$\frac{1}{5}\mathrm{\Omega }$ for $R_4$ and $\frac{1}{5}\mathrm{\Omega }$ for ${\mathrm{R}}_5$ in expression of net resistance.

$\frac{1}{{\mathrm{R}}_{\mathrm{net}}}=\frac{1}{\left({\displaystyle \frac{1}{5}\mathrm{\Omega }}\right)}+\frac{1}{\left({\displaystyle \frac{1}{5}\mathrm{\Omega }}\right)}+\frac{1}{\left({\displaystyle \frac{1}{5}\mathrm{\Omega }}\right)}+\frac{1}{\left({\displaystyle \frac{1}{5}\mathrm{\Omega }}\right)}+\frac{1}{\left({\displaystyle \frac{1}{5}\mathrm{\Omega }}\right)}=\frac{1}{25}\mathrm{\Omega }$

Hence, the Minimum resistance is $\frac{1}{25}\mathrm{\Omega }$.