The error in measuring the current with a tan | vedclass

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Question

(C) In a tangent galvanometer,the current $i$ is given by $i = k 	an \phi$,where $k$ is the reduction factor and $\phi$ is the deflection.Differentia

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) In a tangent galvanometer,the current $i$ is given by $i = k 	an \phi$,where $k$ is the reduction factor and $\phi$ is the deflection.Differentiating both sides with respect to $\phi$,we get $\frac{di}{d\phi} = k \sec^2 \phi$.Thus,the fractional error is given by $\frac{di}{i} = \frac{k \sec^2 \phi \cdot d\phi}{k 	an \phi} = \frac{d\phi}{\sin \phi \cos \phi}$.Multiplying the numerator and denominator by $2$,we get $\frac{di}{i} = \frac{2 d\phi}{2 \sin \phi \cos \phi} = \frac{2 d\phi}{\sin 2\phi}$.For the error $\frac{di}{i}$ to be minimum,$\sin 2\phi$ must be maximum.The maximum value of $\sin 2\phi$ is $1$,which occurs when $2\phi = 90^o$,or $\phi = 45^o$.

The error in measuring the current with a tangent galvanometer is minimum when the deflection is about $45^o$. (in $^o$)

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