Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, ${\rm{100 }}\,{\rm{W,}}\,{\rm{ 60 }}\,{\rm{W}}\,{\rm{ and }}\,{\rm{40}}\,{\rm{ W}}$ bulbs have filament resistances ${R_{100}},\,\,{R_{60}}\,\,{\rm{and}}\,\,{R_{40}}$, respectively, the relation between these resistances is

${R_{100}} = {R_{40}}\,\, + \,\,{R_{60}}$

${R_{100}}\,\,\, > {R_{60}}\,\,\, > {R_{40}}$

$\frac{1}{{{R_{100}}}}\,\,\, > \,\,\,\frac{1}{{{R_{60}}}}\,\,\, > \,\,\frac{1}{{{R_{40}}}}$

JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature,

\mathrm{100 }\,\mathrm{W,}\,\mathrm{ 60 }\,\mathrm{W}\,\mathrm{ and }\,\mathrm{40}\,\mathrm{ W}

bulbs have filament resistances

{R_{100}},\,\,{R_{60}}\,\,\mathrm{and}\,\,{R_{40}}

, respectively, the relation between these resistances is

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Correct option: A

\frac{1}{{{R_{100}}}} = \frac{1}{{{R_{40}}}}\,\, + \,\,\frac{1}{{{R_{60}}}}

Option A (

\frac{1}{{{R_{100}}}} = \frac{1}{{{R_{40}}}}\, + \,\frac{1}{{{R_{60}}}}

) is correct because power is inversely proportional to resistance at constant voltage, and the powers of the bulbs add up.

JEE PHYSICS MCQ: CURRENT ELECTRICITY

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