Current Electricity

Question: The length of a wire of a potentiometer is 100 cm, and the emf of its standard cell is E volt. It is employed to measure the e.m.f of a battery whose internal resistance is 0.5 $\Omega$. If the balance point is obtained at l = 30 cm from the positive end, the e.m.f. of the battery is

1) $\frac{{30E}}{{100}}$

2) $\frac{{30E}}{{100.5}}$

3) $\frac{{30E}}{{(100 - 0.5)}}$

4) $\frac{{30(E - 0.5i)}}{{100}}$, where i is the current in the potentiometer

Solution:

Explanation: No Explanation

Different Measuring Instruments Potentiometer Meter Bridge

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