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(B) To convert a galvanometer into a voltmeter, a high resistance $R_s$ must be connected in series with the galvanometer coil.Given:Galvanometer resi

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$8990$

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(B) To convert a galvanometer into a voltmeter, a high resistance $R_s$ must be connected in series with the galvanometer coil.Given:Galvanometer resistance $R_G = 10 \Omega$Full scale deflection current $I_g = 2 	ext{ mA} = 2 	imes 10^{-3} 	ext{ A}$Required voltage range $V = 18 	ext{ V}$The formula for the series resistance is $V = I_g(R_G + R_s)$.Substituting the values:$18 = 2 	imes 10^{-3} 	imes (10 + R_s)$$18 / (2 	imes 10^{-3}) = 10 + R_s$$9000 = 10 + R_s$$R_s = 9000 - 10 = 8990 \Omega$Thus, a resistance of $8990 \Omega$ must be connected in series.

$A$ galvanometer coil has a resistance of $10 \Omega$ and the meter shows full scale deflection for a current of $2 \text{ mA}$. What resistance needs to be connected to convert the meter into a voltmeter of range $0$ to $18 \text{ V}$ (in $\Omega$)?

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