A cell of emf 4V and internal resistance .5 ohms is connected across a load of resistance: i)7.5 ohms ii)11.5 ohms.
calculate i) the ratio of the differenceas in the emf of the cell and the potential drop across the load, ii)the ratio of the currents in the two cases.

this qustion is wrt potentiometer.
sir i wolud like to get a response as soon as possible.

The potential drop across circuit is given as 

 

V=E-Ir

 

=> V= IR-Ir

 

=> V=I(R-r)

 

Also , I(R+r)=E

 

Therefore, In i) V1=I(7.5-0.5)= 7I

E1=I(7.5+0.5)= 8I

              ii) V2=I(11.5-0.5)=11I

        E2=I(11.5+0.5)= 12I

Therefore V1/V2= 7I/11I=7/11

SImilarly E1/E2= 8I/12I= 2/3

  • 6

 Since in a potentiometer, the driver current remains same whatever be the emf of cell be, therefore ratio of current is 1:1

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