In a moving coil galvanometer, two moving coils $M_{1}$ and $M_{2}$ have the following particulars :
$R_{1}=5 \Omega, N_{1}=15, A_{1}=3.6 \times 10^{-3} m^{2}, B_{1}=0.25 T$
$R_{2}=7 \Omega, N_{2}=21, A_{2}=1.8 \times 10^{-3} \mathrm{~m}^{2}, B_{2}=0.50 \mathrm{~T}$
Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_{1}$ and $M_{2}$ ?
Select the correct option:
A
$1: 2$
B
$1: 3$
C
$1: 1$
D
$1: 4$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Voltage sensitivity $=\frac{\theta}{V}=\frac{N A B}{c R}$
Ratio $=\left(\frac{N_{1} A_{1} B_{1}}{N_{2} A_{2} B_{2}}\right) \frac{R_{2}}{R_{1}}=\frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5} \times \frac{7}{5}=\frac{1}{1}$
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