JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 10)
A galvanometer with its coil resistance 25 $$\Omega $$ requires a current of 1 mA for its full deflection. In order to construct an ammeter to read upto a current of 2 A, the approximate value of the shunt resistance should be :
$$2.5 \times {10^{ - 3}}\,\Omega $$
$$1.25 \times {10^{ - 2}}\Omega $$
$$1.25 \times {10^{ - 3}}\Omega $$
$$2.5 \times {10^{ - 2}}\Omega $$
Explanation
_16th_April_Morning_Slot_en_10_1.png)
Given,
Ig = 1 ma
I $$-$$ Ig = 2 A
Rg = 25 $$\Omega $$
$$\therefore\,\,\,$$ Ig Rg = (I $$-$$ Ig) S
$$ \Rightarrow $$ $$\,\,\,$$ S = $${{{{10}^{ - 3}} \times 25} \over 2}$$
$$ \Rightarrow $$ $$\,\,\,$$ S = 1.25 $$ \times $$ 10$$-$$2 $$\Omega $$
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