A) \[1.0\,\,\Omega \]
B) \[0.5\,\,\Omega \]
C) \[2.0\,\,\Omega \]
D) zero
Correct Answer: A
Solution :
| [a] Key Idea: This problem is based on the application of potentiometer in which we find the internal resistance of a cell. |
| In potentiometer experiment in which we find infernal resistance of a cell, let E be the emf of the cell and V the terminal potential difference, then\[\frac{E}{V}=\frac{{{l}_{1}}}{{{l}_{2}}}\] where \[{{l}_{1}}\] and \[{{l}_{2}}\] are lengths of potentiometer wire with and without short circuited through a resistance. |
| Since, \[\frac{E}{V}=\frac{R+r}{R}\] |
| [\[\because \] \[E=I(R+r)\] and \[V=IR\]] |
| \[\therefore \] \[\frac{R+r}{R}=\frac{{{l}_{1}}}{{{l}_{2}}}\] |
| or \[1+\frac{r}{R}=\frac{110}{100}\] |
| or \[\frac{r}{R}=\frac{10}{100}\] |
| or \[r=\frac{1}{10}\times 10=1.\Omega \] |
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