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A voltmeter of resistance $20000 \Omega$ reads 5 volt. To make it read 20 volt, the extra resistance required is
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Verified Answer
The correct answer is:
$60000 \Omega$ in series
For a voltmeter, in which current $I_g$ passes,
where $G=$ resistance of voltmeter $=20000 \Omega$
$\therefore \quad$ From (i) and (ii)
$\frac{5}{20}=\frac{G}{G+R} \quad$ or $\quad \frac{1}{4}=\frac{G}{G+R}$
or $R=3 G=3 \times 20000=60000$ or extra resistance $=60000 \Omega$ in series
where $G=$ resistance of voltmeter $=20000 \Omega$
$\therefore \quad$ From (i) and (ii)
$\frac{5}{20}=\frac{G}{G+R} \quad$ or $\quad \frac{1}{4}=\frac{G}{G+R}$
or $R=3 G=3 \times 20000=60000$ or extra resistance $=60000 \Omega$ in series
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