The galvanometer resistance is 30 Ω and it is connected to 2 V battery along with a resistance 2000 Ω in series. A full scale deflection of 25 divisions is obtained. In order to reduce this deflection to 20 divisions, the resistance in series should be

2470 Ω

2370 Ω

2180 Ω

2210 Ω

Solution

2470 Ω

Total initial resistance = R_G + R

= 30 + 2000

= 2030Ω

And connected with battery E = 2V

Current i = $\frac{2}{2030}$ A

as the full-scale deflection scale for

1 mA = 25 divisions

Let x be the effective resistance of the circuit, then

2V = 2030 × $\frac{2}{2030}$ × 10³

= x × $\frac{2}{25}$

⇒ x = $\frac{2 \times 25}{2} \times 10^{3}$

⇒ x = 2500 Ω

∴ Resistance to be added= 2500 - 30 = 2470 Ω

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