A galvanometer,whose resistance is 50, Omega, | vedclass

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(B) The full-scale deflection current $(I_g)$ of the galvanometer is calculated by multiplying the current per division by the total number of divisio

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$200$

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(B) The full-scale deflection current $(I_g)$ of the galvanometer is calculated by multiplying the current per division by the total number of divisions:$I_g = (4 	imes 10^{-4}\, A/	ext{division}) 	imes 25\, 	ext{divisions} = 10^{-2}\, A$.To convert a galvanometer into a voltmeter of range $V$,a high resistance $R$ must be connected in series with the galvanometer resistance $G$.The formula is $V = I_g(G + R)$.Given $V = 2.5\, V$,$I_g = 10^{-2}\, A$,and $G = 50\, \Omega$:$2.5 = 10^{-2} 	imes (50 + R)$.Dividing both sides by $10^{-2}$:$250 = 50 + R$.Solving for $R$:$R = 250 - 50 = 200\, \Omega$.

$A$ galvanometer,whose resistance is $50\, \Omega$,has $25$ divisions. When a current of $4 \times 10^{-4}\, A$ passes through it,its needle deflects by one division. To use this galvanometer as a voltmeter of range $2.5\, V$,it should be connected to a resistance of ....... $\Omega$.

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

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