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A current of $500 \mu \mathrm{A}$ deflects a coil of a moving coil galvanometer through 60. What should be the current to cause the rotation through $\frac{\pi}{6}$ radians? What is the current sensitivity of the galvanometer? If resistance of galvanometer is $65\Omega$ , find its voltage sensitivity.
Option: 1
$0.12,2.4 \times 10^3$

Option: 2
$2.5 \times 10^4, 0.12$

Option: 3
$0.14,2.5 \times 10^6$

Option: 4
$0.14,5.2 \times 10^3$

Answers (1)

best_answer

$I_1=500 \mu \mathrm{A}$

$\theta_1 = 60^{\circ}$

$\theta_2=\frac{\pi}{6} \mathrm{rad}=30^{\circ}, I_2=\text { ? }$

For galvanometer, $I_1=\frac{K}{n B A} \theta_1$

$I_2=\frac{K}{n B A} \theta_2$ $\frac{I_2}{I_1}=\frac{\theta_2}{\theta_1}$

$\text { or } \quad I_2=\frac{\theta_2}{\theta_1} \times I_1=\frac{30^{\circ}}{60^{\circ}} \times 500 \mu \mathrm{A}$

$=\frac{1}{2} \times 500=250 \mathrm{\mu A}$

Current sensitivity, $I_S=\frac{\theta_1}{I_1}=\frac{60^{\circ}}{500 \mathrm{~\mu A}}$

$=0.12 \text { degree } / \mu \mathrm{A}$

Voltage sensitivity, $V_s=\frac{I_s}{R}=\frac{0.12 \operatorname{degree} \left(\mu A\right)^{-1}}{50}$

$=\frac{0.12 \times\left(10^{-6}\right)^{-1}}{50 \Omega}=0.0024 \times 10^6$

$=2.4 \times 10^3$ degree/ $\mu A$

Posted by

Rishabh

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