Question

A galvanometer coil has resistance of 30 ohm ans

Answer 1

Answer:

V=IR

V=0.3 * R

Ig = V/G+R

Ig(G+R) = V

R= (V/Ig) - G

R=( 0.3R/2*10-3)-30

By taking LCM

R(2*10-3) = 0.3R - 30* (2 * 10-3)

R(2*10-3-0.3)= -30 * (2*10-3 )

R(-298 * 10-3) = -60 * 10-3

298R = 60

R = 60/298

R = 0.2013 ohms

1. make me brainliest
2. follow me also
3. always ready to give your questions solution

Answer 2

100 Ω

Shunt resistance, S=

I−I

g

I

g

G

where I

g

is the full scale deflection current of the galvanometer, I is the maximum current to be read on the ammeter.

I

g

=0.1I,I

s

=0.9I;S=I

g

R

g

/I

s

=0.1×900/0.9=100Ω