In a moving coil galvanometer, two moving coils and have the following particulars: Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of and ? [2025]

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Solution

Q.
In a moving coil galvanometer, two moving coils $M_{1}$ and $M_{2}$ have the following particulars:

$R_{1} = 5 Ω, N_{1} = 15, A_{1} = 3.6 \times 10^{- 3} m^{2}, B_{1} = 0.25 T$

$R_{2} = 7 Ω, N_{2} = 21, A_{2} = 1.8 \times 10^{- 3} m^{2}, B_{2} = 0.50 T$

Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_{1}$ and $M_{2}$ [2025]

1 1 : 1

2 1 : 4

3 1 : 3

4 1 : 2

Ans.
(1)

$K θ = N I A B \Rightarrow \frac{θ}{I} = \frac{N A B}{K}$

Voltage sensitivity $= \frac{θ}{V} = \frac{θ}{R I} = \frac{N A B}{K R}$

Ratio of voltage sensitivity $= (\frac{N_{1} A_{1} B_{1}}{R_{1}}) (\frac{R_{2}}{N_{2} A_{2} B_{2}})$

Ratio $= (\frac{N_{1} A_{1} B_{1}}{N_{2} A_{2} B_{2}}) \frac{R_{2}}{R_{1}} = \frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5} \times \frac{7}{5} = \frac{1}{1}$

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