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! \[-{{i}_{a}}\] will flow through shunt resistance.                                 Potential difference across ammeter and shunt resistance is same. \[i.e.,{{i}_{a}}\times R=(i-{{i}_{a}})\times S\] \[orS=\frac{{{i}_{a}}R}{i-{{i}_{a}}}....(i)\] Given,   \[{{i}_{a}}=100\,A,\,\,i=750\,A,R=13\,\Omega \]                 Hence,   \[S=\frac{100\times 13}{750-100}=2\,\Omega \]