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  • A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r1. An unknown e.m.f. E is balanced at a leng

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r1. An unknown e.m.f. E is balanced at a length I of the potentiometer wire. The e.m.f. E will be given by:

Option: 1

\frac{{\text{E}_0 \text{r}}} {{\left( {\text{r} + \text{r}_\text{1} } \right)}} \cdot \frac{l} {\text{L}}


Option: 2

\frac{{\text{E}_0 l}} {\text{L}}


Option: 3

\frac{{\text{LE}_0 \text{ r}}} {{\left( {\text{r} + \text{r}_\text{1} } \right)l}}


Option: 4

\frac{{\text{LE}_0 \text{ r}}} {{l\,\text{r}_1 }}


Answers (1)

best_answer

Current in the circuit having E_{0}$ is: $i=\frac{E_{0}}{r+\eta_{1}}

voltage dropped in potentiometer wire of length L is: i r=\frac{E_{n} r}{r+r_{1}}

since the balanced point is obtained at length I of the potentiometer wire, the emf of the cell E has to match the voltage dropped in length I of the wire.


\therefore E=\frac{E_{0}r}{r+r_{1}} \times\left(\frac{l}{L}\right)

Posted by

Deependra Verma

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