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Q. A milliammeter of range $10\, mA$ has a coil of resistance $1\, \Omega$. To use it as a voltmeter of range $10\, V$, the resistance that must be connected in series with it is

KCETKCET 2022

Solution:

$Wkt,\, R=( V / Ig )- G =\left(10 / 10 \times 10^{3}\right)-1=1000-1=999\, \Omega$