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Two cells of emf E1 and E2 are shown,when a potentiometer is connected between A and B the balancing length of the potentiometer wire is 300cm.on connecting same potentiometer between A and C,the balancing length is 100cm.the ratio E1÷E2 is ?

Two cells of emf E1 and E2 are shown,when a potentiometer is connected between A and B the balancing length of the potentiometer wire is 300cm.on connecting same potentiometer between A and C,the balancing length is 100cm.the ratio E1÷E2 is ?

Grade:12

2 Answers

Vikas TU
14149 Points
3 years ago
Dear Student,
​The question is here incomplete.
Pls do check the qstn. and Repost it again for the proper clarification.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Akash rawat
13 Points
3 years ago
See guys according to question E1=k×300For second part it depends on the connection of cell if negative terminal on same side then E1+E2=k×100If in opposite direction thenE1-E2=k×100Hope u got it Akash RawatClass 12(jms)Dehradun

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