Answer
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Hint
Question is asking about the range of ammeters to increase. Range of ammeter is the maximum value of current an ammeter can read. For reading higher values of current, resistance should be minimum.
Complete step by step answer
To convert a galvanometer to ammeter we connect a shunt resistance parallel to the galvanometer. Since, the connection is parallel means the voltage across the shunt and the galvanometer (which is used to make the ammeter) branch is same but the current through both the branches is different. In a broader way since the current I flows through the galvanometer which has an order of mille - amperes and it cannot let more current pass through it and we have to increase the range of current. So, we connect a shunt resistor of negligible resistance so that all the current passes through the shunt branch and the current which passes through the galvanometer branch is ignored. So, as a whole we get the reading from the shunt and this entire circuit is said to be an ammeter.
From the above explanation it is clear that we require negligible resistance, which means that the resistance should be decreased.
Current through galvanometer is,
$\Rightarrow {I_g} = \dfrac{{{R_s}}}{{{R_s} + {R_g}}}I $
$\Rightarrow I = (\dfrac{{{R_s} + {R_g}}}{{{R_s}}}){I_g}$
$\Rightarrow I = (1 + \dfrac{{{R_g}}}{{{R_s}}}){I_g} $
Hence, the correct answer is option (B).
Note
The circuit of the galvanometer with shunt connected in parallel is the ammeter and not only the galvanometer. i.e the entire circuit is called ammeter.
Question is asking about the range of ammeters to increase. Range of ammeter is the maximum value of current an ammeter can read. For reading higher values of current, resistance should be minimum.
Complete step by step answer
To convert a galvanometer to ammeter we connect a shunt resistance parallel to the galvanometer. Since, the connection is parallel means the voltage across the shunt and the galvanometer (which is used to make the ammeter) branch is same but the current through both the branches is different. In a broader way since the current I flows through the galvanometer which has an order of mille - amperes and it cannot let more current pass through it and we have to increase the range of current. So, we connect a shunt resistor of negligible resistance so that all the current passes through the shunt branch and the current which passes through the galvanometer branch is ignored. So, as a whole we get the reading from the shunt and this entire circuit is said to be an ammeter.
From the above explanation it is clear that we require negligible resistance, which means that the resistance should be decreased.
Current through galvanometer is,
$\Rightarrow {I_g} = \dfrac{{{R_s}}}{{{R_s} + {R_g}}}I $
$\Rightarrow I = (\dfrac{{{R_s} + {R_g}}}{{{R_s}}}){I_g}$
$\Rightarrow I = (1 + \dfrac{{{R_g}}}{{{R_s}}}){I_g} $
Hence, the correct answer is option (B).
Note
The circuit of the galvanometer with shunt connected in parallel is the ammeter and not only the galvanometer. i.e the entire circuit is called ammeter.
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