question_answer

A galvanometer of resistance 500 is connected to a battery of 3 V along with a resistance of \[2950\,\Omega \] in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be                                                                                                              [AIPMPT (S) 2008]

A)  \[5050\,\,\Omega \]       

B)       \[5550\,\,\Omega \]

C)  \[6050\,\,\Omega \]       

D)       \[4450\,\,\Omega \]

Correct Answer: D

Solution :

[d] Current through the galvanometer

\[I=\frac{3}{(50+2950)}={{10}^{-3}}A\]

Current for 30 divisions \[={{10}^{-3}}A\]

Current for 20 divisions \[=\frac{{{10}^{-3}}}{30}\times 20\]

\[=\frac{2}{3}\times {{10}^{-3}}A\]

For the same deflection to obtain for 20 divisions, let resistance added be R

\[\therefore \]      \[\frac{2}{3}\times {{10}^{-3}}=\frac{3}{(50+1R)}\]

or         \[R=4450\,\Omega \]