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

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

Correct Answer: B

Solution :

 The potential difference across ammeter and shunt is the 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 the same. That \[{{i}_{a}}\times R=(i-{{i}_{a}})\times S\] or         \[S=\frac{{{i}_{a}}R}{i-{{i}_{a}}}\] Given,\[{{i}_{a}}=1000\,A,i=750\,A,R=13\,\Omega \] Hence,\[\frac{100\times 13}{750}=2\Omega \]