A galvanometer of resistance 50 Ω gives full scale deflection for a current of 0.05 A. Calculate the length of the shunt wire required to convert the galvanometer into an ammeter of range 0 to 5 A. The diameter of the shunt wire is 2 mm and its resistivity is 5 x 10^–7 Ω m.

Solution

Given,
Resistance of galvanometer, R_g = 50 Ω
Current, I_g = 0.05 A
Ammeter current, I = 5A
Diameter of the shunt wire, d = 2 mm = 2

\times

10^-3 m
Therefore, radius, r = 1

\times 10^{- 3} m

Now,
By using the formula,

S = \frac{{GI}_{g}}{I - I_{g}}

Putting the values,

S = \frac{50 \times 0.05}{5 - 0.05} Ω = \frac{50}{99} Ω

Also,

S = p \frac{1}{{πr}^{2}} or l = \frac{{πr}^{2} S}{p}

Putting the values, we get

l = \frac{22}{7} \times {(1 \times 10^{- 3})}^{2} \times \frac{50}{99} \times \frac{1}{5 \times 10^{- 7}} = 3.175 m .

which, is the required length of the shunt wire.

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