question_answer

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

A) \[20\,\Omega \]                       

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

C) \[\text{0}.2\,\,\Omega \]                      

D) \[2\,\,k\Omega \]

Correct Answer: B

Solution :

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