JEE Main 2017 — Current Electricity Question with Solution

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

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

The power is given by

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

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

$$P\propto \frac{1}{R}$$

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

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

Therefore, $$P_2>P_2>P_3$$