Question

A galvanometer of resistance 50 ohm is connected to a battery of 3 v along with resistance of 2950 in series a full scale deflection of 30 division is obtain in the galvanometer . In order to reduce this deflection to 20 division ,the resistance in series should be

Answer 1

Answer:

Total initial resistance

=G+R=50Ω+2950Ω=3000Ω

Current,I=

3000Ω

3V

=1×10

−3

A=1mA

If the deflection has to be reduced to 20 divisions, then current

I

′

=

30

1mA

×20=

3

2

mA

Let x be the effective resistance of the circuit,

3V=3000Ω×1mA=xΩ X

3

2

mA

orx=3000×1×

2

3

=4500Ω

∴ Resistance to be added =(4500Ω−50Ω)=4450Ω