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A galvanometer of resistance 50 Ω is connected to a battery of 8 V along with a resistance of 3950 Ω in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 15 divisions, the resistance in series should be _____.

Asked in GUJCET 2015 · Ammeters and voltmeters in circuits

Answer: (4) 7950 Ω

Step-by-step solution

Given: G = 50 Ω, ε = 8 V, series resistance 3950 Ω; deflection 30 divisions, wanted 15 divisions.

Idea: the deflection is proportional to the current, so half the deflection needs half the current, i.e. twice the total resistance.

Now: I = 8/(50 + 3950) = 2×10⁻³ A for 30 divisions.

For 15 divisions: I' = 1×10⁻³ A, so the total resistance must be 8/(10⁻³) = 8000 Ω.

Series resistance = 8000 - 50 = 7950 Ω.

So the resistance in series should be 7950 Ω.

Why the other options are wrong

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