Question
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.
$$2\,k\Omega $$
Answer :
$$2\,\Omega $$
Solution :
Let $${i_a}$$ be the current flowing through ammeter and $$i$$ be the total current. So, a current $$i - {i_a}$$ will flow through shunt resistance.
Potential difference across ammeter and shunt resistance is same.
i.e. $${i_a} \times R = \left( {i - {i_a}} \right) \times S$$
or $$S = \frac{{{i_a}R}}{{i - {i_a}}}\,.......\left( {\text{i}} \right)$$
Given, $${i_a} = 100\,A,i = 750\,A,R = 13\,\Omega $$
Hence, $$S = \frac{{100 \times 13}}{{750 - 100}} = 2\,\Omega $$
Let $${i_a}$$ be the current flowing through ammeter and $$i$$ be the total current. So, a current $$i - {i_a}$$ will flow through shunt resistance.
Potential difference across ammeter and shunt resistance is same.
i.e. $${i_a} \times R = \left( {i - {i_a}} \right) \times S$$
or $$S = \frac{{{i_a}R}}{{i - {i_a}}}\,.......\left( {\text{i}} \right)$$
Given, $${i_a} = 100\,A,i = 750\,A,R = 13\,\Omega $$
Hence, $$S = \frac{{100 \times 13}}{{750 - 100}} = 2\,\Omega $$
