
When 2 ampere current is passed through a tangent galvanometer, it gives a deflection of 30 degree. For 60 degree deflection, the current must be
Answer
164.1k+ views
Hint: Tangent galvanometers were the early instruments used to measure low electric currents. It is made out of a spiral of insulated copper wire coiled around a nonmagnetic circular frame. Its operation is based upon the principle of magnetism's tangent law. A magnetic field (B) is formed at the centre of the circular coil when a current is carried through it in a perpendicular direction to the surface of the coil.
Complete answer:
Electric current in tangent galvanometer is directly proportional to the angle of deflection i.e. $I\propto \tan \theta $. The equation for the electric current for a tangent galvanometer is $I=k\tan \theta $ , where K is the reduction factor of the tangent galvanometer.
Given, when the current is 2 ampere the angle of deflection is 30, we need to find out the amount of current to be passed through it to get 60 degree of deflection
We know that, $I=k\tan \theta $ $\cdots (1)$
In first case,
${{I}_{1}}=2~amphere$ and
$\theta =30$
Now putting the values in equation (1)
$2=k\tan 30$$\cdots (2)$
In second case, we need to find out ${{I}_{2}}$ for the angle of deflection being 60
${{I}_{2}}=k\tan 60$$\cdots (3)$
Now by dividing equation 2 and equation 3, we will get the value of ${{I}_{2}}$
$\dfrac{2}{{{I}_{2}}}=\dfrac{k\tan 30}{k\tan 60}$
We know that the value of
$\tan 30=\dfrac{1}{\sqrt{3}}$ and value of $\tan 60=\sqrt{3}$
$\dfrac{2}{{{I}_{2}}}=\dfrac{\dfrac{1}{\sqrt{3}}}{\sqrt{3}}$
$\Rightarrow \dfrac{2}{{{I}_{2}}}=\dfrac{1}{3}$
$\Rightarrow {{I}_{2}}=6$
Therefore, the value of current for 60 degree deflection is 6 ampere.
Note:For these kind of questions, we must be familiar with the notion of trigonometry and the value of tan linked with various angles. Furthermore, we must consider the relationship between electric current and angle of deflection.
Complete answer:
Electric current in tangent galvanometer is directly proportional to the angle of deflection i.e. $I\propto \tan \theta $. The equation for the electric current for a tangent galvanometer is $I=k\tan \theta $ , where K is the reduction factor of the tangent galvanometer.
Given, when the current is 2 ampere the angle of deflection is 30, we need to find out the amount of current to be passed through it to get 60 degree of deflection
We know that, $I=k\tan \theta $ $\cdots (1)$
In first case,
${{I}_{1}}=2~amphere$ and
$\theta =30$
Now putting the values in equation (1)
$2=k\tan 30$$\cdots (2)$
In second case, we need to find out ${{I}_{2}}$ for the angle of deflection being 60
${{I}_{2}}=k\tan 60$$\cdots (3)$
Now by dividing equation 2 and equation 3, we will get the value of ${{I}_{2}}$
$\dfrac{2}{{{I}_{2}}}=\dfrac{k\tan 30}{k\tan 60}$
We know that the value of
$\tan 30=\dfrac{1}{\sqrt{3}}$ and value of $\tan 60=\sqrt{3}$
$\dfrac{2}{{{I}_{2}}}=\dfrac{\dfrac{1}{\sqrt{3}}}{\sqrt{3}}$
$\Rightarrow \dfrac{2}{{{I}_{2}}}=\dfrac{1}{3}$
$\Rightarrow {{I}_{2}}=6$
Therefore, the value of current for 60 degree deflection is 6 ampere.
Note:For these kind of questions, we must be familiar with the notion of trigonometry and the value of tan linked with various angles. Furthermore, we must consider the relationship between electric current and angle of deflection.
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