question_answer

The resistance of an ammeter is \[13\Omega \] and its scale is graduated for a current upto\[100\,\,A\]. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto \[750\,\,A\] by this meter. The value of shunt resistance is

A) \[20\,\,\Omega \]                           

B) \[2\,\,\Omega \]

C) \[0.2\,\,\Omega \]                         

D)  \[2\,\,k\Omega \]

Correct Answer: B

Solution :

Key Idea The potential difference across ammeter and shunt is same. Let \[{{i}_{a}}\] is the current flowing through ammeter and \[i\] is the total current. So, a current \[i-{{i}_{a}}\] will flow through shunt resistance. Potential difference across ammeter and shunt resistance is same. \[ie,\]    \[{{i}_{a}}\times R=(i-{{i}_{a}})\times S\] or            \[S=\frac{{{i}_{a}}R}{i-{{i}_{a}}}\]                                              ? (i) Given,   \[{{i}_{a}}=100\,\,A,\,\,i=750\,\,A,\,\,R=13\,\,\Omega \] Hence,  \[S=\frac{100\times 13}{750-100}=2\,\,\Omega \]