A voltmeter of resistance 20000,Omega reads 5 | vedclass

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(C) For a voltmeter,the current $I_g$ passing through it is given by $V = I_g G$,where $G$ is the resistance of the voltmeter.Initially,for a range of

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$60000\,\Omega$ in series

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(C) For a voltmeter,the current $I_g$ passing through it is given by $V = I_g G$,where $G$ is the resistance of the voltmeter.Initially,for a range of $5 	ext{ V}$:$5 = I_g 	imes 20000 \quad \dots(i)$To increase the range to $20 	ext{ V}$,we connect an additional resistance $R$ in series with the voltmeter. The new total resistance becomes $(G + R)$.For the new range of $20 	ext{ V}$:$20 = I_g 	imes (G + R) \quad \dots(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{20}{5} = \frac{I_g(G + R)}{I_g G}$$4 = \frac{G + R}{G}$$4G = G + R$$R = 3G$Given $G = 20000\,\Omega$,we have:$R = 3 	imes 20000 = 60000\,\Omega$Thus,an extra resistance of $60000\,\Omega$ must be connected in series.

$A$ voltmeter of resistance $20000\,\Omega$ reads $5 \text{ V}$. To make it read $20 \text{ V}$,the extra resistance required is

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