Question

A galvanometer of resistance 10 $\Omega$ gives full - scale deflection when 1 mA current passes through it. The resistance required to convert it into a voltmeter reading upto 2.5 V is

• A. $24.9 \Omega$
• B. $249 \Omega$
• C. 2490$\Omega$
• D. $24900 \Omega$

Answer 1

Answer: (C)

2490$\Omega$

Answer 2

Here

$\mathrm { G } = 10 \Omega , \mathrm { V } = 2.5 \mathrm {~V}$

From the figure

$\mathrm { V } = \mathrm { I } _ { \mathrm { g } } ( \mathrm { G } + \mathrm { R } )$ or $\mathrm { R } = \frac { \mathrm { V } } { \mathrm { I } _ { \mathrm { g } } } - \mathrm { G }$

Substituting the given values we get

$\mathrm { R } = \frac { 2.5 \mathrm {~V} } { 1 \times 10 ^ { - 3 } \mathrm {~A} } - 10 \Omega = 2500 \Omega - 10 \Omega = 2490 \Omega$