(C) The full-scale deflection current $I_g$ is calculated by multiplying the number of divisions by the current per division: $I_g = 50 \times 2 \time

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The full-scale deflection current $I_g$ is calculated by multiplying the number of divisions by the current per division: $I_g = 50 \times 2 \times 10^{-4} \text{ A} = 10^{-2} \text{ A}$.To convert a galvanometer into a voltmeter,a high resistance $R_H$ is connected in series with it.The formula for the series resistance is $R_H = \frac{V}{I_g} - R_g$,where $V = 25 \text{ V}$ and $R_g = 25 \Omega$.Substituting the values: $R_H = \frac{25}{10^{-2}} - 25 = 2500 - 25 = 2475 \Omega$.

Class 12 Physics Moving Charges and Magnetism The Moving Coil Galvanometer (Sensitivity) and Ammeter and Voltmeter Conversion

Medium

$A$ galvanometer with $50$ divisions on the scale has a resistance of $25 \Omega$. $A$ current of $2 \times 10^{-4} \text{ A}$ gives a deflection of one scale division. The additional series resistance required to convert it into a voltmeter reading up to $25 \text{ V}$ is $.... \Omega$.

A
$1200$
B
$1225$
C
$2475$
D
$2500$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Moving Charges and Magnetism

Subtopic

The Moving Coil Galvanometer (Sensitivity) and Ammeter and Voltmeter Conversion

We have a galvanometer of resistance $25\,\Omega$. It is shunted by a $2.5\,\Omega$ wire. The part of the total current that flows through the galvanometer is given as:

Medium

View Solution

If only $2 \%$ of the total current passes through an ammeter having a coil of resistance $R$,then the resistance of the shunt of the ammeter is:

MediumMHT CET 2020

View Solution

An ammeter with a resistance of $1\, \Omega$ can measure up to $10\, mA$. To convert it into a voltmeter that can measure up to $10\, V$,what resistance (in $\Omega$) must be connected in series?

Easy

View Solution

$A$ galvanometer of resistance $36 \ \Omega$ is converted into an ammeter by using a shunt of $4 \ \Omega$. The fraction $f_0$ of the total current passing through the galvanometer is:

Medium

View Solution

Give the uses of a galvanometer.

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ galvanometer with $50$ divisions on the scale has a resistance of $25 \Omega$. $A$ current of $2 \times 10^{-4} \text{ A}$ gives a deflection of one scale division. The additional series resistance required to convert it into a voltmeter reading up to $25 \text{ V}$ is $.... \Omega$.

Similar Questions

We have a galvanometer of resistance $25\,\Omega$. It is shunted by a $2.5\,\Omega$ wire. The part of the total current that flows through the galvanometer is given as:

If only $2 \%$ of the total current passes through an ammeter having a coil of resistance $R$,then the resistance of the shunt of the ammeter is:

An ammeter with a resistance of $1\, \Omega$ can measure up to $10\, mA$. To convert it into a voltmeter that can measure up to $10\, V$,what resistance (in $\Omega$) must be connected in series?

$A$ galvanometer of resistance $36 \ \Omega$ is converted into an ammeter by using a shunt of $4 \ \Omega$. The fraction $f_0$ of the total current passing through the galvanometer is:

Give the uses of a galvanometer.

Vedclass Test Series

Exam Paper Generator

Online Exam Module