Question

in a tangent galvanometer a current of 1 ampere produces deflection of 30 degree the current required to produce a deflection of 60°​

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

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 ${\bf \red{3\:A}}$.

Answer 2

Answer:

Iαtanθ

\implies{\sf \dfrac{i_1}{i_2}=\dfrac{tan\theta_1}{tan\theta_2} }⟹

i

2

i

1

=

tanθ

2

tanθ

1

\implies{\sf \dfrac{1}{i_2}=\dfrac{tan30°}{tan60°}}⟹

i

2

1

=

tan60°

tan30°

tan30° = {\sf \dfrac{1}{\sqrt{3}}}

3

1

tan60° = {\sf \sqrt{3}}

3

\implies{\sf \dfrac{1}{i_2}=\dfrac{1/\sqrt{3}}{\sqrt{3}}}⟹

i

2

1

=

3

1/

3

\implies{\sf \dfrac{1}{i_2}=\dfrac{1}{3}}⟹

i

2

1

=

3

1

\implies{\bf \red{ i_2=3\:A}}⟹i

2

=3A

\huge\underline{\underline{\bf \orange{Answer-}}}

Answer−

Current required to produce a deflection of 60° is {\bf \red{3\:A}}3A .