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, \[{{i}_{a}}=100A,i=750A,R=13\,\Omega \] Hence, \[S=\frac{100\times 13}{750-100}=2\,\Omega \]