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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

\(\mathrm{A}\) bar magnet of length \(10 \mathrm{~cm}\) and having the pole strength equal \(10^{3} \mathrm{Am}\) to is kept in a magnetic field having magnetic induction (B) equal to \(4 \pi \times 10^{3}\) tesla. It makes an angle of \(30^{\circ}\) with the direction of magnetic induction. The value of the torque acting on the magnet is Joule. (a) \(2 \pi \times 10^{-7}\) (b) \(2 \pi \times 10^{5}\) (c) \(0.5\) (d) \(0.5 \times 10^{2}\)

The deflection in a Galvanometer falls from 50 division to 20 when \(12 \Omega\) shunt is applied. The Galvanometer resistance is (a) \(18 \Omega\) (b) \(36 \Omega\) (c) \(24 \Omega\) (d) \(30 \Omega\)

The forces existing between two parallel current carrying conductors is \(\mathrm{F}\). If the current in each conductor is doubled, then the value of force will be (a) \(2 \mathrm{~F}\) (b) \(4 \mathrm{~F}\) (c) \(5 \mathrm{~F}\) (d) \((\mathrm{F} / 2)\)

The unit of ele. current "AMPERE" is the current which when flowing through each of two parallel wires spaced 1 meter apart in vacuum and of infinite length will give rise to a force between them equal to $\mathrm{N} / \mathrm{m}$ (a) 1 (b) \(2 \times 10^{-7}\) (c) \(1 \times 10^{-2}\) (d) \(4 \pi \times 10^{-7}\)

An element \(\mathrm{d} \ell^{-}=\mathrm{dx} \uparrow\) (where $\mathrm{dx}=1 \mathrm{~cm}$ ) is placed at the origin and carries a large current \(\mathrm{I}=10 \mathrm{Amp}\). What is the mag. field on the Y-axis at a distance of \(0.5\) meter ? (a) \(2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (b) \(4 \times 10^{8} \mathrm{k} \wedge \mathrm{T}\) (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (d) \(-4 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\)
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