When 2 A current is passed through a tangent | vedclass

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(D) The current $i$ in a tangent galvanometer is related to the deflection $\phi$ by the formula $i = K 	an \phi$,where $K$ is the reduction factor o

Vedclass · Accepted Answer

$6$

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(D) The current $i$ in a tangent galvanometer is related to the deflection $\phi$ by the formula $i = K 	an \phi$,where $K$ is the reduction factor of the galvanometer.Therefore,$i \propto 	an \phi$.Given $i_1 = 2 \ A$ and $\phi_1 = 30^\circ$.We need to find $i_2$ for $\phi_2 = 60^\circ$.Using the ratio: $\frac{i_1}{i_2} = \frac{	an \phi_1}{	an \phi_2}$.Substituting the values: $\frac{2}{i_2} = \frac{	an 30^\circ}{	an 60^\circ}$.Since $	an 30^\circ = \frac{1}{\sqrt{3}}$ and $	an 60^\circ = \sqrt{3}$,we have $\frac{2}{i_2} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$.Thus,$i_2 = 2 	imes 3 = 6 \ A$.

When $2 \ A$ current is passed through a tangent galvanometer,it gives a deflection of $30^\circ$. For $60^\circ$ deflection,the current must be........$A$.

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