Current through the galvanometer I=(50+2950)3=10−3 A Current for 30 divisions =10−3 A Current for 20 divisions =3010−3×20 =32×10−3 A For the same deflection to obtain for 20 divisions, let resistance added be R ∴32×10−3=(50+1R)3or R=4450Ω
A galvanometer of resistance 50 Ω is connected to a battery of 3 ~V along with a resistance of 2950 Ω in series. A full scale deflection of 30…
NEET UG 2008 — Chemistry Physical Chemistry
2008mcqmedium
A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance of 2950Ω 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
Official previous-year question
Held on 30 Apr 2008 · Verified 9 Jul 2026.
Options
- A
5050Ω
- B
5550Ω
- C
6050Ω
- D
4450Ω
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