Question

A galvanometer, whose resistance is $50\text{ohm}$, has $25$ divisions in it. When a current of $4\times {10}^{-4}\text{A}$ passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range $2.5\text{V}$, it should be connected to a resistance of :

• A. $\text{6250 ohm}$
• B. $\text{250 ohm}$
• C. $\text{6200 ohm}$
• D. $\text{200 ohm}$

Answer 1

The correct option is D $\text{200 ohm}$

Galvanometer has $25$ divisions, therefore the maximum deflection will be shown for the current

$I_g=4\times {10}^{-4}\times 25={10}^{-2}\text{A}$


From the above circuit,

$V=I_g(G+R)$
$2.5=(50+R){10}^{-2}$
$\therefore R=200\Omega$

Hence, option $\left(B\right)$ is correct.