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Grade 12Electric Current

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 ?

Pujitha

9 Years agoGrade 12

2 Answers

Vikas TU

9 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. 4^th year
Thapar University)

Akash rawat

8 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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