Exam

AIIMS

Year

2016

Subject

Physics

Chapter

Current Electricity

AIIMS 2016 · Physics · Current Electricity

A voltmeter of resistance 20000 reads 5 volt. To make it read 20...

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

Step-by-step solution

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