The length of a wire of a potentiometer is 100 cm and the e.m.f. of its standard cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is If the balance point is obtained at l=30 cm from the positive end, the e.m.f. of the battery is :

 

  • Option 1)

    \frac{30E}{100.5}

  • Option 2)

    \frac{30E}{100-0.5}

  • Option 3)

    \frac{30E}{100}-0.5 i         where is the current in the potentiometer wire

  • Option 4)

    \frac{30E}{100}

 

Answers (1)

As we learnt in

Potential gradient -

V=iR= \left (\frac{e}{R+R_{n}+r} \right )R

x=\frac{V}{L}=\frac{e}{(R+R_{n}+r)}\frac{R}{L}

- wherein

r- internal resistance

 

 

Equation of cell -

E=V+iR

- wherein

(E>V)

 

 

 

Potential gradient along wire = \frac{E}{100}

K= \frac{E}{100}\frac{v}{cm}

For battery V=E-ir

E = E.m.f. of battery

K\times 30= E-ir        Where current i is drawn from battery.

\frac{E}{100}\times30=E-ir

\frac{E}{100}\times30=E+ir

\Rightarrow \frac{E}{100}\times300=E+0.5i

E=\frac{30E}{100}-0.5i

 

 


Option 1)

\frac{30E}{100.5}

This option is incorrect.

Option 2)

\frac{30E}{100-0.5}

This option is incorrect.

Option 3)

\frac{30E}{100}-0.5 i         where is the current in the potentiometer wire

This option is correct.

Option 4)

\frac{30E}{100}

This option is incorrect.

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