QJEEM17Ph6c27: JEE Main 2017 Physics - Class 12 JEEM17Ph6c27 Solution: The figure shows three circuits I, II and III which are…

We have the following cases :

$\bullet$ For configuration (I) : Considering as Wheatstone's bridge, we get the equivalent resistance as ${({R_I})_{eq}} = 1\,\Omega$ .

$\bullet$ For configuration (II) : The given configuration is further reduced as follows:

That is, ${({R_{II}})_{eq}} = {1 \over 2}\,\Omega$ .

$\bullet$ For configuration (III) : Similarly, we get ${({R_{III}})_{eq}} = 2\,\Omega$ .

The power is given by

$P = {{{V^2}} \over R}$

Since V is the same for all three configurations, we have the power as a quantity indirectly proportional to the resistance :

$P \propto {1 \over R}$

From the above three case, we have ${({R_I})_{eq}} = 1\,\Omega ;\,{({R_{II}})_{eq}} = {1 \over 2}\,\Omega ;\,{({R_{III}})_{eq}} = 2\,\Omega$ , which implies that

{R_{II}} > {R_I} > {R_{III}}

Therefore, {P_2} > {P_2} > {P_3}