Question

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

Answer 1

A galvanometer of resistance 50 Ω is connected to a battery along with a resistance 2950 Ω in series.

so, Req = $R_G+R_1$

= 50 + 2950 = 3000 Ω

now, current through galvanometer, I = 3v/3000 = 10^-3 A

current for 30 divisions = I = 10^-3 A

⇒current for 1 division = I/30 = 1/30 × 10^-3 A

⇒current for 20 divisions = 20I/30 = 20/30 × 10^-3 A = 2 × 10^-3/3 A

now, equivalent resistance will be , Req = 3v/(2 × 10^-3/3) = 9000/2 = 4500 Ω

as it is given, resistance of galvanometer is 50 Ω

so, resistance in series should be , R = Req - $R_G$

= 4500 - 50 = 4450 Ω

Answer 2

Answer:

4450 ohm

Explanation:

here divisions mean current as galvanometer detects current in circuits

so

I=V/R and Req=2950+50=3000

30=3/3000 divided by 20=3/x

solving it we get 4500

now 50 ohm is resistance of galvanometer

so required R=4500-50=4450