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The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its standard cell is volt. It is employed to measure the e.m.f. of a battery whose internal resistance is 0.5Ω. If the balance point is obtained at = 30 cm from the positive end, e.m.f. of the battery is Where is the current in the potentiometer wire.

Option: 1

\mathrm{\frac{30 E}{100}}


Option: 2

\mathrm{\frac{30 E}{100.5}}


Option: 3

\mathrm{\frac{30 E}{(100-0.5)}}


Option: 4

\mathrm{\frac{30(E-0.5 i)}{100}}


Answers (1)

best_answer

Using the principle of potentiometer, \mathrm{V \propto l}
\mathrm{ \therefore \frac{V}{E}=\frac{l}{L} \text { or } V=\frac{l}{L} E=\frac{30}{100} E=\frac{30 E}{100} }

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rishi.raj

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