The equivalent resistance
Question:

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)

Solution:

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

$\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\mathrm{p}$…(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 $\mathrm{n}$

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

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

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