A galvanometer having 30 divisions has a curr | vedclass

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(B) $1$. First,calculate the full-scale deflection current $(I_g)$ of the galvanometer.$I_g = \frac{	ext{Total divisions}}{	ext{Current sensitivity}

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$12.5 	ext{ k}\Omega$

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(B) $1$. First,calculate the full-scale deflection current $(I_g)$ of the galvanometer.$I_g = \frac{	ext{Total divisions}}{	ext{Current sensitivity}} = \frac{30 	ext{ div}}{0.0625 	ext{ div}/\mu A} = 480 \mu A = 480 	imes 10^{-6} 	ext{ A} = 4.8 	imes 10^{-4} 	ext{ A}$.$2$. $A$ voltmeter is formed by connecting a high resistance $(R)$ in series with the galvanometer.$3$. The total resistance of the voltmeter $(R_v)$ is given by the formula $V = I_g 	imes R_v$,where $V$ is the maximum voltage to be measured.$4$. Rearranging for $R_v$: $R_v = \frac{V}{I_g} = \frac{6 	ext{ V}}{4.8 	imes 10^{-4} 	ext{ A}}$.$5$. $R_v = \frac{6}{4.8} 	imes 10^4 \Omega = 1.25 	imes 10^4 \Omega = 12.5 	ext{ k}\Omega$.

$A$ galvanometer having $30$ divisions has a current sensitivity of $0.0625 \frac{\text{div}}{\mu A}$. If it is converted into a voltmeter to read a maximum of $6 \text{ V}$,then the resistance of that voltmeter is:

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