
The sensitiveness of tangent galvanometer will be maximum if deflection in it is tending to
A. $0{}^\circ$
B. ${{30}^{{}^\circ }}$
C. $45{}^\circ$
D. ${{60}^{{}^\circ }}$
Answer
233.1k+ views
Hint: We first need to calculate the error when a small deflection is taken into consideration in the galvanometer. Therefore, we can calculate the change in current per angle change in deflection. From this, finally we can calculate the maximum deflection.
Complete step by step answer:
Let us consider $\mathrm{d} \theta$ as the small error while measuring the deflection $\theta$. This is taken so that the corresponding error in the current calculated is di while measuring a current 'i'. We can calculate i as:
$\mathrm{i}=\mathrm{k} \tan \theta \ldots \ldots \ldots .(1)$
While differentiating both side of the derived equation with respect to $\theta$
$\dfrac{di}{d\theta }=k{{\sec }^{2}}\theta$
$\therefore di=k{{\sec }^{2}}\theta \cdot d\theta \ldots \ldots .(2)$
Dividing equation ( 2 ) by ( 1 ), we get,
$\dfrac{\text{d}i}{i}=\dfrac{k\cdot {{\sec }^{2}}\theta \cdot \text{d}\theta }{k\cdot \tan \theta}\dfrac{\text{d}\theta }{{{\cos }^{2}}\theta \cdot \dfrac{\sin \theta }{\cos \theta }}$
$\Rightarrow \dfrac{\text{d}i}{i}=\dfrac{\text{d}\theta }{\cos \theta \cdot \sin \theta }=\dfrac{2\text{d}\theta }{2\cos \theta \cdot \sin \theta }$
$\Rightarrow \dfrac{\text{di}}{i}=\dfrac{2\text{d}\theta }{\sin 2\theta }$
Hence, it is clear to us that the accuracy is maximum when the error is minimum. This means that the quantity $\mathrm{di} / \mathrm{i}$ is minimum when the quantity $(\sin 2 \theta)$ is
Maximum. Therefore, we can calculate,
$\sin 2\theta =1$
$\Rightarrow \quad 2\theta ={{\sin }^{-1}}(1)$
$\Rightarrow \quad 2\theta ={{90}^{{}^\circ }}$
$\Rightarrow \quad \theta ={{45}^{{}^\circ }}$
Therefore, we get that the sensitiveness of a tangent galvanometer is maximum when the
Deflection in it is tending towards ${{45}^{{}^\circ }}$.
Note: The accuracy of the measurement of deflection can be increased by using pointers which are lighter and longer, also increasing the radius of each turn in the coil will increase the accuracy of measurement. Decreasing the number of turns in the coil will also increase the accuracy of measurement.
Complete step by step answer:
Let us consider $\mathrm{d} \theta$ as the small error while measuring the deflection $\theta$. This is taken so that the corresponding error in the current calculated is di while measuring a current 'i'. We can calculate i as:
$\mathrm{i}=\mathrm{k} \tan \theta \ldots \ldots \ldots .(1)$
While differentiating both side of the derived equation with respect to $\theta$
$\dfrac{di}{d\theta }=k{{\sec }^{2}}\theta$
$\therefore di=k{{\sec }^{2}}\theta \cdot d\theta \ldots \ldots .(2)$
Dividing equation ( 2 ) by ( 1 ), we get,
$\dfrac{\text{d}i}{i}=\dfrac{k\cdot {{\sec }^{2}}\theta \cdot \text{d}\theta }{k\cdot \tan \theta}\dfrac{\text{d}\theta }{{{\cos }^{2}}\theta \cdot \dfrac{\sin \theta }{\cos \theta }}$
$\Rightarrow \dfrac{\text{d}i}{i}=\dfrac{\text{d}\theta }{\cos \theta \cdot \sin \theta }=\dfrac{2\text{d}\theta }{2\cos \theta \cdot \sin \theta }$
$\Rightarrow \dfrac{\text{di}}{i}=\dfrac{2\text{d}\theta }{\sin 2\theta }$
Hence, it is clear to us that the accuracy is maximum when the error is minimum. This means that the quantity $\mathrm{di} / \mathrm{i}$ is minimum when the quantity $(\sin 2 \theta)$ is
Maximum. Therefore, we can calculate,
$\sin 2\theta =1$
$\Rightarrow \quad 2\theta ={{\sin }^{-1}}(1)$
$\Rightarrow \quad 2\theta ={{90}^{{}^\circ }}$
$\Rightarrow \quad \theta ={{45}^{{}^\circ }}$
Therefore, we get that the sensitiveness of a tangent galvanometer is maximum when the
Deflection in it is tending towards ${{45}^{{}^\circ }}$.
Note: The accuracy of the measurement of deflection can be increased by using pointers which are lighter and longer, also increasing the radius of each turn in the coil will increase the accuracy of measurement. Decreasing the number of turns in the coil will also increase the accuracy of measurement.
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