Electric current (I) is measured by a galvano | vedclass

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(B) Given that $I \propto 	an 	heta$, we can write $I = k 	an 	heta$. Taking the derivative with respect to $	heta$, we get $dI = k \sec^2 	heta

Vedclass · Accepted Answer

$\frac{\pi}{2} \%$

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(B) Given that $I \propto 	an 	heta$, we can write $I = k 	an 	heta$. Taking the derivative with respect to $	heta$, we get $dI = k \sec^2 	heta \, d	heta$. The relative error is given by $\frac{dI}{I} = \frac{k \sec^2 	heta \, d	heta}{k 	an 	heta} = \frac{\sec^2 	heta}{	an 	heta} d	heta$. Using $\sec^2 	heta = \frac{1}{\cos^2 	heta}$ and $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$, we have $\frac{dI}{I} = \frac{1}{\sin 	heta \cos 	heta} d	heta = \frac{2}{\sin(2	heta)} d	heta$. Given $	heta = 45^{\circ} = \frac{\pi}{4}$ radians and the percentage error in $	heta$ is $\frac{d	heta}{	heta} 	imes 100 = 1\%$, so $d	heta = \frac{1}{100} 	imes \frac{\pi}{4} = \frac{\pi}{400}$ radians. Substituting the values: $\frac{dI}{I} 	imes 100 = \frac{2}{\sin(2 	imes 45^{\circ})} 	imes d	heta 	imes 100 = \frac{2}{\sin(90^{\circ})} 	imes \frac{\pi}{400} 	imes 100 = 2 	imes 1 	imes \frac{\pi}{4} = \frac{\pi}{2} \%$. Thus, the percentage error in the current is $\frac{\pi}{2} \%$.

Electric current $(I)$ is measured by a galvanometer, where the current is proportional to the tangent of the angle $( \theta)$ of deflection. If the deflection is read as $45^{\circ}$ and an error of $1\%$ is made in reading the angle, the percentage error in the current is:

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