Question

What is the value of the shunt resistance
that allows 40% of the main current
through the galvanometer of 1002?​

Answer 1

Answer:

Hello mate ⬇️

Explanation:

Given :   I

g

=5mA=0.005A           G=100Ω              I=1A

Let a shunt resistance R

s

is connected, as shown in the figure, to convert a galvanometer into an ammeter.

From figure,   V

AB

=V

CD

∴    (I−I

g

)R

s

=I

g

G

OR    (1−0.005)R

s

=0.005×100

OR   0.995R

s

=0.5             ⟹R

s

=

9.95

5

Ω