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One filament takes 10min to heat a kettle and another takes 15 min. If connected in parallel they combindly take………min to heat the same kettle:A. 6B. 12.5C. 25D. 7.5 |
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Answer» Correct Answer - A For first filament `H=(V^(2))/(R_(1)) t_(1)rArr R_(1)=(V^(2))/H t_(1).....(i)` For second filament `H=(V^(2))/(R_(2))t_(2)rArr R_(2)=(V^(2))/H T_(2).....(ii)` When place din parallel `H=(V^(2))/(R_(P)) t_(2) rArr R_(p)=(V^(2))/H t_(p)....(iii)` From eqs. (i),(ii),(iii) we get `1/(R_(P))=1/(R_(1))+1/(R_(2))` `rArr H/(V^(2)t_(p))=H/(V^(2)t_(1))+H/(V^(2)t_(2))` `rArr 1/(t_(p)) =1/10 +1/15 rArr t_(p)` =6 min |
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