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Question

To send 10% of the main current through a moving coil galvanometer of resistance 99 the shunt required is-


A
11
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B
9
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C
100
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D
19
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Solution

The correct option is A 11


The current through galvanometer is $$\dfrac{I}{10}.$$

Now, current  through galvanometer is 

$$\dfrac{I}{10}=\dfrac{R}{R+R_g}I$$

$$\implies \dfrac{1}{10}=\dfrac{R}{R+99}$$

$$\implies 10R=R+99$$

$$\implies 9R=99\Omega$$

$$\implies R=11\Omega$$

Answer-(A)

845735_581843_ans_8b6dae259cc4483eada3e2d0486f97b3.png

Physics

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