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