question_answer

The resistance of an ammeter is 13\[\Omega \] and its scale is graduated for a current upro 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)         2k\[\Omega \]

Correct Answer: B

Solution :

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. i.e.,        \[{{i}_{a}}\times R=(i={{i}_{a}})\times S\] or                   \[S=\frac{{{i}_{a}}R}{i-{{i}_{a}}}\]                       ?(i)        Given, \[{{i}_{a}}=100A,\,i=750A,\,R=13\,\Omega \] Hence, \[S=\frac{100\times 13}{750-100}=2\,\Omega \]