QJEEAdv08Phff95: JEE Advanced 2008 Physics - Class 12 JEEAdv08Phff95 Solution: Figure shows three resistor configurations R1, R2 and R3…

We know that $P = {{{V^2}} \over R}$

If potential is constant we have

$P \propto {1 \over R}$

Case I

Hence, this is a clear case of wheat stone bridge R $_1$ = 1 $\Omega$

Case II

Equivalent resistance (R $_2$ )

$\therefore$ ${1 \over {{R_2}}} = {1 \over 2} + {1 \over 1} + {1 \over 2} = {{1 + 2 + 1} \over 2} = {4 \over 2} = 2\,\Omega$

Hence, ${R_2} = {1 \over 2} = 0.5\,\Omega$

It is clear that the equivalent resistance (R $_2$ ) will be less than 1 $\Omega$ .

Case III

Hence, R $_3$ = 2 $\Omega$

Since, R $_2$ < R $_1$ < R $_2$

$\therefore$ P $_2$ > P $_1$ > P $_3$ ( $\because$ $P \propto {1 \over R}$ )