# The equivalent resistance of series combination of two resistors

Question:

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

(Round off to the Nearest Integer)

Solution:

(4)

$\mathrm{R}_{1}+\mathrm{R}_{2}=\mathrm{s} \quad \ldots(1)$

$\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\mathrm{p} \ldots$

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

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

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

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

for minimum value of $\mathrm{n}$

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

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