in a tangent galvanometer a current of 1 ampere produces deflection of

1.	in a tangent galvanometer a current of 1 ampere produces deflection of 30 degree the current required to produce a deflection of 60°
Answer» $\huge\underline{\underline{\bf \orange{Question-}}}$ In a tangent galvanometer a CURRENT of 1 ampere produces deflection of 30 degree the current required to PRODUCE a deflection of 60° $\huge\underline{\underline{\bf \orange{Solution-}}}$ $\large\underline{\underline{\sf Given:}}$ Current ${\sf (i_1)}$ = 1A Deflection ${(\theta_1)}$ = 30° Deflection ${(\theta)}$ = 60° $\large\underline{\underline{\sf To\:Find:}}$ Current required to produce deflection of 60° ${\sf (i_2)}$ . For tangent galvanometer relationship between Current and deflection angle is given by ⎯ ✪ $\large{\boxed{\bf \blue{I=ktan\theta} }}$ Here, K = CONSTANT ⠀⠀⠀⠀i = Current ⠀⠀⠀⠀ ${\theta}$ = angle of deflection We can say that $\large{\bf \blue{I\:\:\alpha\:\:tan\theta} }$ $\implies{\sf \dfrac{i_1}{i_2}=\dfrac{tan\theta_1}{tan\theta_2} }$ $\implies{\sf \dfrac{1}{i_2}=\dfrac{tan30°}{tan60°}}$ tan30° = ${\sf \dfrac{1}{\sqrt{3}}}$ tan60° = ${\sf \sqrt{3}}$ $\implies{\sf \dfrac{1}{i_2}=\dfrac{1/\sqrt{3}}{\sqrt{3}}}$ $\implies{\sf \dfrac{1}{i_2}=\dfrac{1}{3}}$ $\implies{\bf \red{ i_2=3\:A}}$ $\huge\underline{\underline{\bf \orange{Answer-}}}$ Current required to produce a deflection of 60° is ${\bf \red{3\:A}}$ .