The length of a potentiometer wire is 1200 cm and it carries a current of 80mA .FOR A CELL F EMF 4 V AND INTERNAL RESISTANCE 20 ohm the null point is found to be a 1000cm .If a voltmeter is connected across the cell the balancing length is decreased by 20 cm.Find i) resistance of the whole wire
     ii) reading of the voltmeter
      iii)resistance of voltmeter

Asked by Subhashree Lenka | 14th May, 2015, 07:38: PM

Expert Answer:

Given that the length of the potentiometer wire = 1200 cm
Current through the potentiometer wire, I = 80 mA = 80 × 10-3 A
1000 cm of potentiometer wire balances the cell of emf 4 V
 
begin mathsize 14px style left parenthesis straight i right parenthesis Therefore space the space resistance space of space the space 1000 space cm space of space wire comma straight R apostrophe equals fraction numerator 4 over denominator 80 cross times 10 to the power of negative 3 end exponent end fraction equals 50 space straight capital omega The space resistance space of space the space whole space length space of space potentiometer space wire comma space straight R space equals fraction numerator straight R apostrophe over denominator 1000 end fraction cross times 1200 straight R space equals fraction numerator 50 space straight capital omega over denominator 1000 end fraction cross times 1200 straight R space equals 60 space straight capital omega left parenthesis ii right parenthesis When space the space volmeter space is space connected comma space the space balance space length space of space the space potentiometer space will space be colon open parentheses 1000 minus 20 close parentheses cm space equals 980 space cm The space reading space of space the space voltmeter space equals 4 over 1000 cross times 980 space equals 3.92 space straight V left parenthesis iii right parenthesis Let space straight R subscript straight V space be space the space resistance space of space the space voltmeter comma Then space we space have space that colon straight r space equals open parentheses straight E over straight V minus 1 close parentheses straight R subscript straight V straight r space equals open parentheses straight l subscript 1 over straight l subscript 2 minus 1 close parentheses straight R subscript straight V 20 equals open parentheses 1000 over 980 minus 1 close parentheses straight R subscript straight V straight R subscript straight V equals 20 cross times 980 over 20 straight R subscript straight V equals 980 space straight capital omega  end style 

Answered by Jyothi Nair | 15th May, 2015, 09:12: AM