A galvanometer of resistance 25 Ω shows a deflection of 5 divisions when a current of 2 mA is passed through it. If a shunt of 4 Ω is connected and there are 20 divisions on the scale, then the range of the galvanometer is

1 A

58 A

58 mA

30 mA

Solution

58 mA

Given: - Galvanometer resistance R_g = 25 Ω

There are 20 divisions on scale.

Initially, current for 20 divisions

i = $\frac{2 \times 20}{5}$

i = 8 mA

∴ i_g = 8 mA

Let I be the maximum current that galvanometer can read.

$∵ \frac{I_{g}}{I} = \frac{R_{S}}{R_{S} + R_{g}}$

⇒ I= $(\frac{R_{S} + R_{g}}{R_{S}}) I_{g}$

⇒ = $(\frac{4 + 25}{4}) \times 8$

⇒ = 29 × 2

⇒ I = 58 mA

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