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Chapter 13: Problem 1927

The deflection in moving coil Galvanometer is reduced to half when it is shunted with a \(40 \Omega\) coil. The resistance of the Galvanometer is (c) \(40 \Omega\) (a) \(60 \Omega\) (b) \(\overline{10 \Omega}\) (d) \(20 \Omega\)

Short Answer

Expert verified
The resistance of the galvanometer is \(20\Omega\). The correct answer is (d) \(20\Omega\).

Step by step solution

01

Understand the concept of shunt resistance and total resistance

A shunt is a low resistance connected in parallel with a high resistance. In this case, the 40Ω coil is the shunt resistance, and it is connected in parallel with the galvanometer resistance (which we need to find). The total resistance for two resistances in parallel is given by the formula: \[ \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} \]
02

Apply current division formula to find the proportion of current

It is given that the deflection in the moving coil galvanometer is reduced to half when it is shunted with a 40Ω coil, which means the current, when the galvanometer is shunted, is half of the original current. For two resistances in parallel, the current is divided inversely proportionally to the resistance values: \[ \frac{I_1}{I_2} = \frac{R_2}{R_1} \] Since the deflection in the galvanometer is reduced to half, we can write the equation as: \[ \frac{I}{\frac{1}{2}I} = \frac{R_{shunt}}{R_{galv}} \] Where \(R_{shunt} = 40\Omega\) is the shunt resistance and \(R_{galv}\) is the resistance of the galvanometer.
03

Solve for the resistance of the galvanometer

Simplify the proportion equation to find the value of \(R_{galv}\) \[\frac{1}{\frac{1}{2}} = \frac{40\Omega}{R_{galv}}\] \[2 = \frac{40\Omega}{R_{galv}}\] Now, we can solve for \(R_{galv}\): \[ R_{galv} = \frac{40\Omega}{2} \] \[ R_{galv} = 20\Omega \] The resistance of the galvanometer is \(20\Omega\). The correct answer is (d) \(20\Omega\).

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Most popular questions from this chapter

A bar magnet having a magnetic moment of \(2 \times 10^{4} \mathrm{JT}^{-1}\) is free to rotate in a horizontal plane. A horizontal magnetic field \(\mathrm{B}=6 \times 10^{-4}\) Tesla exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction \(60^{\circ}\) from the field is (a) \(0.6 \mathrm{~J}\) (b) \(12 \mathrm{~J}\) (c) \(6 \mathrm{~J}\) (d) \(2 \mathrm{~J}\)

The angles of dip at two places are \(30^{\circ}\) and \(45^{\circ}\). The ratio of horizontal components of earth's magnetic field at the two places will be (a) \(\sqrt{3}: \sqrt{2}\) (b) \(1: \sqrt{2}\) (c) \(1: 2\) (d) \(1: \sqrt{3}\)

1: A charged particle moves in a uniform mag. field. The velocity of the particle at some instant makes an acute angle with the mag. field. The path of the particle will be (a) A straight line (b) A circle (c) A helix with uniform pitch (d) A helix with non-uniform pitch

A conducting circular loop of radius a carries a constant current I. It is placed in a uniform magnetic field \(\mathrm{B}^{-}\), such that \(\mathrm{B}^{-}\) is perpendicular to the plane of the Loop. The magnetic force acting on the Loop is (a) \(\mathrm{B}^{-} \operatorname{Ir}\) (b) \(\mathrm{B}^{-} \mathrm{I} \pi \mathrm{r}^{2}\) (c) Zero (d) BI \((2 \pi \mathrm{r})\)

0: Due to 10 Amp of current flowing in a circular coil of \(10 \mathrm{~cm}\) radius, the mag. field produced at its centre is \(\pi \times 10^{-3}\) Tesla. The number of turns in the coil will be (a) 5000 (b) 100 (c) 50 (d) 25
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