JEE MAIN - Physics (2005 - No. 7)
A moving coil galvanometer has $$150$$ equal divisions. Its current sensitivity is $$10$$- divisions per milliampere and voltage sensitivity is $$2$$ divisions per millivolt. In order that each division reads $$1$$ volt, the resistance in $$ohms$$ needed to be connected in series with the coil will be -
$${10^5}$$
$${10^3}$$
$$9995$$
$$99995$$
Explanation
KEY CONCEPT : Resistance of Galvanometer,
$$G = {{Current\,\,\,sensitivity} \over {Voltage\,\,\,sensityvity}} \Rightarrow G = {{10} \over 2} = 5\Omega $$
Here $${i_g} = $$ Full scale deflection current $$ = {{150} \over {10}} = 15\,\,mA$$
$$V=$$ voltage to be measured $$=150$$ volts
(such that each division reads $$1$$ volt)
$$ \Rightarrow R = {{150} \over {15 \times {{10}^{ - 3}}}} - 5 = 9995\Omega $$
$$G = {{Current\,\,\,sensitivity} \over {Voltage\,\,\,sensityvity}} \Rightarrow G = {{10} \over 2} = 5\Omega $$
Here $${i_g} = $$ Full scale deflection current $$ = {{150} \over {10}} = 15\,\,mA$$
$$V=$$ voltage to be measured $$=150$$ volts
(such that each division reads $$1$$ volt)
$$ \Rightarrow R = {{150} \over {15 \times {{10}^{ - 3}}}} - 5 = 9995\Omega $$
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