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)         \[2k\,\Omega \]

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