Question

What is the lowest total resistance that can be secured by combinations of four coils of resistance $4\mathrm{\Omega },8\mathrm{\Omega },12\mathrm{\Omega },24\mathrm{\Omega }$?

Answer 1

Step 1: Given information

Combinations of four coils of resistance are:

${\mathrm{R}}_1=4\mathrm{\Omega }$

${\mathrm{R}}_2=8\mathrm{\Omega }$

${\mathrm{R}}_3=12\mathrm{\Omega }$

${\mathrm{R}}_4=24\mathrm{\Omega }$

Step 2: Calculation

If these coils are connected in parallel, the equivalent resistance, given by, will be the lowest. Net resistance is given by

$$\begin{gathered} \frac{1}{{\mathrm{R}}_{\mathrm{net}}}=\frac{1}{4\mathrm{\Omega }}+\frac{1}{8\mathrm{\Omega }}+\frac{1}{12\mathrm{\Omega }}+\frac{1}{24\mathrm{\Omega }} \\ \frac{1}{{\mathrm{R}}_{\mathrm{net}}}=\frac{6+3+2+1}{24\mathrm{\Omega }} \\ \frac{1}{{\mathrm{R}}_{\mathrm{net}}}=\frac{12}{24\mathrm{\Omega }} \\ \frac{1}{{\mathrm{R}}_{\mathrm{net}}}=\frac{1}{2\mathrm{\Omega }} \\ {\mathrm{R}}_{\mathrm{net}}=2\mathrm{\Omega } \end{gathered}$$

Hence, the lowest resistance is $2\mathrm{\Omega }$.