question_answer

The resistance of a galvanometer is \[50\Omega \] and the current required to give full scale deflection is \[100\mu A\]. In order to convert into an ammeter for reading up to 10 A, it is necessary to put a resistance of

A) \[5\times {{10}^{-5}}\Omega \]      

B)       \[5\times {{10}^{-3}}\Omega \]

C)        \[5\times {{10}^{-2}}\Omega \]        

D)        \[5\times {{10}^{-4}}\Omega \]

Correct Answer: D

Solution :

                The resistance of the galvanometer \[G=50\Omega \] The current required for a full scale deflection \[{{I}_{FS}}=100\mu A\] \[\therefore \] The voltage applied across the galvanometer \[E=50\times 100\times {{10}^{-6}}\] \[=5\times {{10}^{-3}}V=5mV\] Now to convert this galvanometer into an ammeter we need to put a shunt S in parallel with the galvanometer. The total resistance will be determined by the maximum current read by the ammeter which is 10 A \[\therefore \]  \[R=\frac{E}{I}=\frac{5\times {{10}^{-3}}}{10}=5\times {{10}^{-4}}\Omega \] We can find S from the relation \[\frac{1}{R}=\frac{1}{S}+\frac{1}{G}\] \[\Rightarrow \]  \[\frac{1}{R}=\frac{1}{S}-\frac{1}{G}=\frac{1}{5\times {{10}^{-4}}}-\frac{1}{50}\] \[=\frac{{{10}^{5}}}{50}-\frac{1}{50}\overset{\sim }{\mathop{=}}\,\frac{{{10}^{5}}}{50}\] \[\therefore \]  \[S=\frac{50}{{{10}^{5}}}=5\times {{10}^{-4}}\Omega \]