A galvanometer of resistance 25 Ω is connected to a battery of 2 volt along with a resistance in series. When the value of this resistance is 3000 Ω, a full scale deflection of 30 units is obtained in the galvanometer. In order to reduce this deflection to 20 units, the resistance in series will be
The deflection in a galvanometer is inversely proportional to the series resistance. Reducing the deflection from 30 to 20 units means increasing the resistance by a factor of 30/20 = 3/2. Therefore, the new resistance is
3000 × 3/2 = 4500, plus the galvanometer resistance (25 Ω), totaling approximately 4514 Ω. Hence, the correct answer is (A) 4514 Ω.
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Given:
Galvanometer resistance (G) = 25 Ω
Battery voltage (V) = 2 V
Initial series resistance (R) = 3000 Ω
Full-scale deflection = 30 units
New deflection = 20 units
The current sensitivity is proportional to deflection:
I₂/I₁ = 20/30 = 2/3
Current through the circuit:
I₁ = V/G+R = 2/3000+25
New resistance:
I₂ = 2/G+R’
2/3/1 = G+R/G+R’
R’ = 4514Ω
Answer: (A) 4514 Ω
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