Question

A galvanometer whose resistance is $50\Omega$ has $25$ divisions in it . When a current 0f $4\times {10}^{-4}\text{A}$ passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range $5\text{V}$, it should be connected to a resistance of

• A. $450\Omega \text{in parallel}$
• B. $200\Omega \text{in series}$
• C. $450\Omega \text{in series}$
• D. $200\Omega \text{in parellel}$

Answer 1

The correct option is C $450\Omega \text{in series}$

$I_g=4\times {10}^{-4}\times 25={10}^{-2}\text{A};G=50\Omega ;R=?$


Since, $V=I_g(G+R)$

Substituting the data given in the question, $5={10}^{-2}\times (50+R)$

$\Rightarrow 500=50+R$

$\Rightarrow =450\Omega \text{in series with galvanometer}$

Hence, option (c) is the correct answer.