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Chapter 13: Problem 1921

A Galvanometer has a resistance \(\mathrm{G}\) and \(\mathrm{Q}\) current \(\mathrm{I}_{\mathrm{G}}\) flowing in it produces full scale deflection. \(\mathrm{S}_{1}\) is the value of the shunt which converts it into an ammeter of range 0 to \(\mathrm{I}\) and \(\mathrm{S}_{2}\) is the value of the shunt for the range 0 to \(2 \mathrm{I}\). The ratio $\left(\mathrm{S}_{1} / \mathrm{S}_{2}\right) \mathrm{is}$ (a) $\left[\left(2 \mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) /\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right)\right]$ (b) $(1 / 2)\left[\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) /\left(2 \mathrm{I}-\mathrm{I}_{\mathrm{G}}\right)\right]$ (c) 2 (d) 1

Short Answer

Expert verified
The short answer to the question is: The ratio \(\frac{S_{1}}{S_{2}}\) is (a) \(\left[\left(2I - I_{G}\right) /\left(I - I_{G}\right)\right]\).

Step by step solution

01

Write down the current distribution for both ranges

In both cases, the current gets distributed between the Galvanometer and the shunt resistances in parallel connection. For range 0 to I, we have: \(I = I_{G} + I_{1}\) For the range 0 to 2I, we have: \(2I = I_{G} + I_{2}\) Here, \(I_{1}\) and \(I_{2}\) are the currents passing through the shunt resistances \(S_{1}\) and \(S_{2}\) respectively.
02

Express shunt resistances using Ohm's Law

Since the Galvanometer and shunt resistances are in parallel, they have the same voltage across them. Using Ohm's Law, we can express \(S_{1}\) and \(S_{2}\) as: \(S_{1} = \frac{V}{I_{1}}\) \(S_{2} = \frac{V}{I_{2}}\) Where V is the voltage across the Galvanometer and the shunt resistances.
03

Find the ratio S1 / S2

To find the ratio of the shunt resistances, we can divide the expressions for S1 and S2: \(\frac{S_{1}}{S_{2}} = \frac{\frac{V}{I_{1}}}{\frac{V}{I_{2}}}\) The voltage V cancels out, and we are left with: \(\frac{S_{1}}{S_{2}} = \frac{I_{2}}{I_{1}}\) Now, from Step 1, we can substitute \(I_{1} = I - I_{G}\) and \(I_{2} = 2I - I_{G}\) in the above expression: \(\frac{S_{1}}{S_{2}} = \frac{2I - I_{G}}{I - I_{G}}\) Therefore, the ratio \(\frac{S_{1}}{S_{2}}\) is (a) \(\left[\left(2I - I_{G}\right) /\left(I - I_{G}\right)\right]\).

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Most popular questions from this chapter

Two similar coils are kept mutually perpendicular such that their centers co- inside. At the centre, find the ratio of the mag. field due to one coil and the resultant magnetic field by both coils, if the same current is flown. (a) \(1: \sqrt{2}\) (b) \(1: 2\) (c) \(2: 1\) (d) \(\sqrt{3}: 1\)

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