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  • A cell of E.M.F. (E) has a resistance connected across it, and a voltmeter measures the potential dif

A cell of E.M.F. (E) has a resistance \mathrm{R} connected across it, and a voltmeter measures the potential difference across the terminals of the cell, which is \mathrm{V}. What is the internal resistance of the cell?
 

Option: 1

\mathrm{2(E-V) \frac{V}{R}}
 


Option: 2

\mathrm{\quad \frac{2(E-v)}{v} R}

 


Option: 3

\mathrm{\frac{(E-V)}{V} R}

 


Option: 4

\mathrm{(E-V) R}


Answers (1)

best_answer

We know that the current flow through in a cell-

\mathrm{I=\frac{E}{r+R}} \leftarrow  External resistance of cell

        \uparrowInternal resistance of cell
but by ohm's law - \mathrm{V=I R, I=\frac{V}{R}}

Now, \mathrm{\frac{V}{R}=\frac{E}{r+R}}

\mathrm{V(r+R)=ER}

\mathrm{Vr+V R=E R }

\mathrm{Vr=ER-VR}

\mathrm{V r=R(E-V) }

\mathrm{r=\frac{(E-V)}{V} \cdot R}

Hence option 3 is correct.







 

Posted by

Divya Prakash Singh

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