Question 402: Current Electricity

Question

A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R = 2400 $$\Omega $$. Deflection becomes 20 divisions when resistance taken from resistance box is 4900 $$\Omega $$. Then we can conclude :

Accepted Answer

Let full scale deflection of current = I
In case 1, when R = 2400 $$\Omega $$ and deflection of 40 divisions present.
$$ 	herefore $$   $${{40} \over {50}}{m I} = {V \over {G + R}}$$
$$ \Rightarrow $$   $${4 \over 5}{m I}$$  = $${2 \over {G + 2400}}$$          . . .(1)
Incase 2, when R = 4900 $$\Omega $$ and deflection of 20 divisions present
$$ 	herefore $$   $${{20} \over {50}}{m I} = {V \over {G + R}}$$
$$ \Rightarrow $$   $${2 \over 5}{m I}$$ = $${2 \over {G + 4900}}$$          . . .(2)
From (1) and (2) we get,
$${4 \over 2} = {{G + 4900} \over {G + 2400}}$$
$$ \Rightarrow $$   2G + 4800 $$=$$ G + 4900
$$ \Rightarrow $$   G $$=$$  100 $$\Omega $$
Putting value of G in equation (1), we get,
$${4 \over 5}{m I} = {2 \over {100 + 2400}}$$
$$ \Rightarrow $$   $${m I} = 1$$ mA
Current sensitivity  $$=$$ $${{m I} \over {number\,\,of\,\,divisions}}$$
$$=$$ $${1 \over {50}}$$
$$=$$ 0.02 mA / division
$$=$$ 20 $$\mu $$A / division
Resistance required for deflection of 10 divisions
$${{10} \over {50}}{m I} = {V \over {G + R}}$$
$$ \Rightarrow $$   $${1 \over 5} 	imes 1 	imes {10^{ - 3}} = {2 \over {100 + R}}$$
$$ \Rightarrow $$   R $$=$$ 9900  $$\Omega $$

Answer

Resistance of galvanometer is 200 $$\Omega $$

Answer

Full scale deflection current is 2 mA.

Answer

Current sensitivity of galvanometer is 20 $$\mu $$A/division.

Answer

Resistance required on R.B. for a deflection of 10 divisions is 9800 $$\Omega $$.

Current Electricity

A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R = 2400 $\Omega . Deflection becomes 20 divisions when resistance taken from resistance box is 4900 \Omega $. Then we can conclude :

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