The equivalent resistance


The equivalent resistance of series combination of two resistors is ' $s$ '. When they are connected in parallel, the equivalent resistance is ' $p$ '. If $s=n p$, then the minimum value for $n$ is

(Round off to the Nearest Integer)



$\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\mathrm{p}$...(2)

$\mathrm{R}_{1} \mathrm{R}_{2}=\mathrm{sp}$

$\mathrm{R}_{1} \mathrm{R}_{2}=\mathrm{np}^{2}$

$\mathrm{R}_{1}+\mathrm{R}_{2}=\frac{\mathrm{nR}_{1} \mathrm{R}_{2}}{\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)}$

$\frac{\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)^{2}}{\mathrm{R}_{1} \mathrm{R}_{2}}=\mathrm{n}$

for minimum value of $\mathrm{n}$


$\therefore \mathrm{n}=\frac{(2 \mathrm{R})^{2}}{\mathrm{R}^{2}}=4$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now