A galvanometer of resistance 20 Omega gives a | vedclass

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(B) Given: Resistance of galvanometer $G = 20 \Omega$. Full scale deflection current $I_g$: Since $5$ divisions correspond to $1 	ext{ mA}$,$50$ divi

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$2480 \Omega$ in series.

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(B) Given: Resistance of galvanometer $G = 20 \Omega$. Full scale deflection current $I_g$: Since $5$ divisions correspond to $1 	ext{ mA}$,$50$ divisions correspond to $I_g = (1 	ext{ mA} / 5) 	imes 50 = 10 	ext{ mA} = 0.01 	ext{ A}$. Required voltage range $V = 25 	ext{ V}$. To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with it. The formula is $V = I_g(R + G)$. Rearranging for $R$: $R = (V / I_g) - G$. Substituting the values: $R = (25 / 0.01) - 20 = 2500 - 20 = 2480 \Omega$. Thus,we must connect $2480 \Omega$ in series.

$A$ galvanometer of resistance $20 \Omega$ gives a deflection of $5$ divisions when $1 \text{ mA}$ current flows through it. The galvanometer scale has $50$ divisions. To convert the galvanometer into a voltmeter of range $25 \text{ V}$,we should connect a resistance of

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