JEE MAIN - Physics (2002 - No. 4)
A wire when connected to $$220$$ $$V$$ mains supply has power dissipation $${P_1}.$$ Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $${P_2}.$$ Then $${P_2}:{P_1}$$ is
$$1$$
$$4$$
$$2$$
$$3$$
Explanation
Case 1 : $${P_1} = {{{V^2}} \over R}$$
Case 2 : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is $${R \over 2}.$$ These are connected in parallel
$$\therefore$$ $${{\mathop{\rm R}\nolimits} _{eq}} = {{R/2} \over 2} = {R \over 4}$$
$$\therefore$$ $${P_2} = {{{V^2}} \over {R/4}} = 4\left( {{{{V^2}} \over R}} \right) = 4{P_1}$$
Case 2 : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is $${R \over 2}.$$ These are connected in parallel
$$\therefore$$ $${{\mathop{\rm R}\nolimits} _{eq}} = {{R/2} \over 2} = {R \over 4}$$
$$\therefore$$ $${P_2} = {{{V^2}} \over {R/4}} = 4\left( {{{{V^2}} \over R}} \right) = 4{P_1}$$
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