A galvanometer has a resistance of 50Ω and it allows maximum current of 5 mA.

Question

A galvanometer has a resistance of \(50\,\Omega\) and it allows maximum current of 5 mA. It can be converted into voltmeter to measure upto 100 V by connecting in series a resistor of resistance 

(1) \(5975\,\Omega\)

(2) \(20050\,\Omega\)

(3) \(19950\,\Omega\)

(4) \(19500\,\Omega\)

Answer 1

(3) \(19950\,\Omega\)

\(R=\frac {V}{I_g}-R_g=\frac {100}{5\times 10^{-3}}-50\)

= 20000 - 50

\(=19950\,\Omega\)