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Asked by Anshu Chauhan 6th March 2015, 10:01 PM
Answered by Expert
Answer:
begin mathsize 14px style Given space that colon Length space of space the space potentiometer space wire comma space straight L space equals 400 space cm Resistance space of space the space potentiometer space wire comma equals space 10 straight capital omega Balancing space point space obtained comma straight l space equals 240 space cm The space current space flowing space through space the space potentiometer space wire comma space straight I space equals straight V over straight R equals fraction numerator 3 space straight V over denominator open parentheses 10 plus 290 close parentheses end fraction equals 0.01 straight A The space potential space unkown space potential space straight V space of space the space circuit colon straight V equals Potential space drop space per space unit space length space of space the space potentimeter space wire cross times Balancing space length The space potential space drop space per space unit space length space of space the space potentimeter space wire space equals fraction numerator 0.01 space straight A space cross times 10 over denominator 400 end fraction equals 2.5 cross times 10 to the power of negative 4 end exponent space straight V divided by cm Given space balancing space length space equals space 240 space cm Therefore colon straight V equals 2.5 cross times 10 to the power of negative 4 end exponent space straight V divided by cm cross times 240 space cm space straight V equals 600 space cross times 10 to the power of negative 4 end exponent space straight V space space equals 60 space mV end style
Answered by Expert 7th March 2015, 10:36 PM
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