question_answer

The resistance of an ammeter is x\[13\,\,\Omega \] 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:                                                                                                                      [AIPMT (S) 2007]

A)  \[20\,\,\Omega \]

B)                   \[2\,\,\Omega \]

C)  \[0.2\,\,\Omega \]

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

Correct Answer: B

Solution :

[b] 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! \[-\text{ }{{i}_{a}}\] will flow through shunt resistance.

Potential difference across ammeter and shunt resistance is same.

i.e.,       \[{{i}_{a}}\times R=(i-{{i}_{a}})\times S\]

or         \[S=\frac{{{i}_{a}}R}{i-{{i}_{a}}}\]                 (i)

Given,   \[{{i}_{a}}=100\,A,\,\,i=750\,A,R=13\,\Omega \]

Hence,   \[S=\frac{100\times 13}{750-100}=2\,\Omega \]