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ICSE Class 10 Answered

Two, cells of emf 1.5v and internal resistance 1.5ohm are connected in parallel.The arrangement is connected to a 2.5ohm resistance.Find potential difference accross the terminals of the cell.
Asked by bhilave | 13 Jun, 2015, 08:45: PM
answered-by-expert Expert Answer
begin mathsize 12px style Two comma space cells space of space emf space 1.5 space straight V space and space internal space resistance space 1.5 space straight capital omega space are space connected space in space parallel. Total space emf comma space straight E equals 1.5 space straight V straight r subscript straight p space is space the space total space internal space resistance space of space the space combination space of space cell space then comma 1 over straight r subscript straight p equals fraction numerator 1 over denominator 1.5 end fraction plus fraction numerator 1 over denominator 1.5 end fraction straight r subscript straight p equals 0.75 space straight capital omega The space arrangement space is space connected space to space straight a space 2.5 space straight capital omega space resistance. Terminal space potential space difference space is space straight V comma we space know space that comma straight r subscript straight p equals fraction numerator straight E minus straight V over denominator straight V end fraction straight R 0.75 equals fraction numerator 1.5 minus straight V over denominator straight V end fraction cross times 2.5 fraction numerator 0.75 straight V over denominator 2.5 end fraction equals 1.5 minus straight V 0.3 straight V equals 1.5 minus straight V 1.3 straight V equals 1.5 straight V equals 1.15 space Volt Hence comma space potential space difference space accross space the space terminals space of space the space cell space is space 1.15 space Volt. end style
Answered by Priyanka Kumbhar | 14 Jun, 2015, 07:20: PM

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