Question

A galvanometer of resistance $5ohms$ gives a full scale deflection for a potential difference of $10mV$. To convert the galvanometer into a voltmeter giving a full scale deflection for a potential difference of $1V,$ the size of the resistance that must be attached to the voltmeter is

• A. $0.495ohm$
• B. $49.5ohm$
• C. $495ohm$
• D. $4950ohm$

Answer 1

Answer: (C)

$495ohm$

$I_g=\frac{10\times {10}^{-3}}{5}$
$=2\times {10}^{-3}A$;

$R=\frac{V}{I_g}-R_g$
$=\frac{1}{2\times {10}^{-3}}-5=495\Omega$