Question

A parallel combination of an $8\Omega$resistor and an unknown resistor R is connected in series with a $16\Omega$resistor and a battery. This circuit is then disassembled and the three resistors are connected in series with each other and the same battery. If in both arrangements, the current through the $8\Omega$resistor is the same, then R is (Take $\sqrt{2}=1.41$) and write upto two digits after the decimal point.)

Answer 1

$V=\left(24+R\right)i---\left(1\right)$$V=\left(\frac{8R}{8+R}+16\right)i\text{'}Buti\text{'}=\left(\frac{8+R}{R}\right)i\left(Voltagesacrossresistorsinparallelareequal\right)\Rightarrow V=\left(\frac{8R}{8+R}+16\right)\left(\frac{8+R}{R}\right)i---\left(2\right)$
Equating (1) and (2),
$\left(24+R\right)i=\left(\frac{8R}{8+R}+16\right)\left(\frac{8+R}{R}\right)i\Rightarrow 24+R=8+16\left(\frac{8+R}{R}\right)\Rightarrow 24R+R^2=24R+128\Rightarrow R^2=128\Rightarrow R=\sqrt{128}=8\sqrt{2}=11.28\Omega$