Current Electricity 4 Question 3
3. A galvanometer whose resistance is $50 \Omega$, has 25 divisions in it. When a current of $4 \times 10^{-4} A$ passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range $2.5 V$, it should be connected to a resistance of
(Main 2019, 12 Jan II)
(a) $250 \Omega$
(b) $6200 \Omega$
(c) $200 \Omega$
(d) $6250 \Omega$
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Solution:
- Current for deflection of pointer by
1 division $=4 \times 10^{-4} A$
So, current for full-scale deflection $=I _g$ $=$ Number of divisions $\times$ Current for 1 division $\Rightarrow \quad I _g=25 \times 4 \times 10^{-4}=1 \times 10^{-2} A$
Now, let a resistance of $R$ is put in series with galvanometer to make it a voltmeter of range $2.5 V$.
Then,
$$ \begin{aligned} I _g(R+G) & =V _{A B} \\ 1 \times 10^{-2}(R+50) & =2.5 \end{aligned} $$
$\left[\because\right.$ Given, $\left.G=50 \Omega, V _{A B}=2.5 V\right]$
$\therefore \quad R=250-50=200 \Omega$
