question_answer

The resistance of an ammeter is 130 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\[\Omega \]

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