Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 1926

A Galvanometer coil has a resistance of \(15 \Omega\) and gives full scale deflection for a current of \(4 \mathrm{~mA}\). To convert it to an ammeter of range 0 to \(6 \mathrm{Amp}\) (a) \(10 \mathrm{~m} \Omega\) resistance is to be connected in parallel to the galvanometer. (b) \(10 \mathrm{~m} \Omega\) resistance is to be connected in series with the galvanometer. (c) \(0.1 \Omega\) resistance is to be connected in parallel to the galvanometer. (d) \(0.1 \Omega\) resistance is to be connected in series with the galvanometer.

Short Answer

Expert verified
To convert the given galvanometer into an ammeter with a range of 0 to 6 A, a \(10 \mathrm{~m} \Omega\) resistance should be connected in parallel to the galvanometer. So, the correct option is (a).

Step by step solution

01

Identify given data

We are given: - Resistance of Galvanometer coil (Rg) = 15 Ω - Full scale deflection current (Ig) = 4 mA = 0.004 A - The desired range of the ammeter = 0 to 6 A Step 2: Calculate the resistance of the shunt resistor (Rs)
02

Calculate the resistance of the shunt resistor (Rs)

Since the galvanometer is to be converted into an ammeter, we need to connect a shunt resistor (Rs) either in parallel or series to achieve the desired range of 6 A. The resistance of the shunt resistor can be found by using the formula: \(R_s = \frac{R_g \cdot I_g}{I - I_g}\) We will substitute our given values and find Rs for each option. (a) Is Rs = 10 mΩ and connected in parallel to the galvanometer?
03

Calculate for option (a)

Let's calculate Rs for a current range of 0 to 6A using the formula above: \(R_s = \frac{15 \Omega \cdot 0.004 A}{6 A - 0.004 A} = 10 \times 10^{-3} \Omega \) The calculated resistance for the shunt resistor is 10 mΩ, and it needs to be connected in parallel to the galvanometer. So option (a) is correct. (b) Is Rs = 10 mΩ and connected in series to the galvanometer?
04

Check for option (b)

Series connection won't work for this case, as it would not create the desired ammeter range. Hence, option (b) is incorrect. (c) Is Rs = 0.1 Ω and connected in parallel to the galvanometer?
05

Check for option (c)

We have already found the correct shunt resistor value in option (a). So option (c) is incorrect. (d) Is Rs = 0.1 Ω and connected in series to the galvanometer?
06

Check for option (d)

Again, a series connection is not suitable for this case. So option (d) is incorrect. #Answer# (a) 10 mΩ resistance is to be connected in parallel to the galvanometer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Force between two identical bar magnets whose centers are I meter apart is \(4.8 \mathrm{~N}\), when their axes are in the same line. If separation is increased to \(2 r\), the force between them is reduced to (a) \(2.4 \mathrm{~N}\) (b) \(1.2 \mathrm{~N}\) (c) \(0.6 \mathrm{~N}\) (d) \(0.3 \mathrm{~N}\)

In each of the following questions, Match column-I and column-II and select the correct match out of the four given choices.\begin{tabular}{l|r} Column-I & Column - II \end{tabular} (A) Magnetic field induction due to Current 1 through straight conductor at a perpendicular distance \(\mathrm{r}\). (B) Magnetic field induction at (Q) \(\left[\left(\mu_{0} \mathrm{I}\right) /(4 \pi \mathrm{r})\right]\) the centre of current \((1)\) carrying Loop of radius (r) (C) Magnetic field induction at the (R) \(\left[\mu_{0} /(4 \pi)\right](2 \mathrm{I} / \mathrm{r})\) axis of current (1) carrying coil of radius (r) at a distance (r) from centre of coil (D) Magnetic field induction at the (S) \(\left[\mu_{0} /(4 \sqrt{2})\right](\mathrm{L} / \mathrm{r})\) at the centre due to circular arc of length \(\mathrm{r}\) and radius (r) carrying current (I) (a) $\mathrm{A} \rightarrow \mathrm{R} ; \mathrm{B} \rightarrow \mathrm{S} ; \mathrm{C} \rightarrow \mathrm{P} ; \mathrm{D} \rightarrow \mathrm{Q}$ (b) $\mathrm{A} \rightarrow \mathrm{R} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{Q}$ (c) $\mathrm{A} \rightarrow \mathrm{P} ; \mathrm{B} \rightarrow \mathrm{Q} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}$ (d) $\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}$

If a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be (a) Only inside the pipe (b) Only outside the pipe (c) Neither inside nor outside the pipe (d) Both inside and outside the pipe

Two straight long conductors \(\mathrm{AOB}\) and \(\mathrm{COD}\) are perpendicular to each other and carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). The magnitude of the mag. field at a point " \(\mathrm{P}^{n}\) at a distance " \(\mathrm{a}^{\prime \prime}\) from the point "O" in a direction perpendicular to the plane \(\mathrm{ABCD}\) is (a) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}+I_{2}\right)\) (b) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}-I_{2}\right)\) (c) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}\right)^{1 / 2}$ (d) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left[\left(I_{1} I_{2}\right) /\left(I_{1}-I_{2}\right)\right]$

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) \(2 \mathrm{~J}\)
See all solutions

Recommended explanations on English Textbooks

Essay Plan

Read Explanation

Language and Social Groups

Read Explanation

Multiple Choice Questions

Read Explanation

Lexis and Semantics Summary

Read Explanation

Summary Text

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free