JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 3)
A moving coil galvanometer allows a full scale
current of 10–4 A. A series resistance of 2 M$$\Omega $$
is required to convert the above galvanometer
into a voltmeter of range 0-5 V. Therefore the
value of shunt resistance required to convert the
above galvanometer into an ammeter of range
0.10 mA is :
200 $$\Omega $$
500 $$\Omega $$
100 $$\Omega $$
None of the options are correct
Explanation
Given data,
$$ \begin{aligned} I =10^{-4} \mathrm{~A}, \\\\ R_S =2 \mathrm{M} \Omega=2 \times 10^6 \Omega, \\\\ V_{\max } =5 \mathrm{~V} \end{aligned} $$
Let internal resistance of galvanometer is $R_G$.
_10th_April_Morning_Slot_en_3_1.png)
Then,
$$ \begin{array}{lc} I \times R_S+I \times R_G=V_{\max } \\\\ \Rightarrow 2 \times 10^6 \times 10^{-4}+10^{-4} \times R_G=5 \\\\ \Rightarrow 10^{-4} R_G=5-200=-195 \\\\ \text { or } R_G=-195 \times 10^4 \Omega \end{array} $$
Resistance cannot be negative.
$\therefore$ No option is correct.
$$ \begin{aligned} I =10^{-4} \mathrm{~A}, \\\\ R_S =2 \mathrm{M} \Omega=2 \times 10^6 \Omega, \\\\ V_{\max } =5 \mathrm{~V} \end{aligned} $$
Let internal resistance of galvanometer is $R_G$.
Then,
$$ \begin{array}{lc} I \times R_S+I \times R_G=V_{\max } \\\\ \Rightarrow 2 \times 10^6 \times 10^{-4}+10^{-4} \times R_G=5 \\\\ \Rightarrow 10^{-4} R_G=5-200=-195 \\\\ \text { or } R_G=-195 \times 10^4 \Omega \end{array} $$
Resistance cannot be negative.
$\therefore$ No option is correct.
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