Question

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

Answer 1

Answer:

The current through galvanometer is

10

I

.

Now, current through galvanometer is

10

I

=

R+R

g

R

I

⟹

10

1

=

R+99

R

⟹10R=R+99

⟹9R=99Ω

Answer 2

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