In an experiment to find the resistance of a galvanometer by the half-deflection method, a $5 V$ battery and a high resistance of $4.9 k\Omega$ are connected in the circuit. In the absence of any shunt resistance, the galvanometer reads $20$ divisions when current flows in the circuit. To reduce the deflection by half, the value of the shunt resistance used is $98 \Omega$. The figure of merit of the galvanometer is given as $..... \mu A / \text{division}$.

  • A
    $5$
  • B
    $20$
  • C
    $25$
  • D
    $50$

Explore More

Similar Questions

$A$ galvanometer can be converted to a voltmeter of full-scale deflection $V_{0}$ by connecting a series resistance $R_{1}$ and can be converted to an ammeter of full-scale deflection $I_{0}$ by connecting a shunt resistance $R_{2}$. What is the current flowing through the galvanometer at its full-scale deflection?

The current passing through the galvanometer is $4\%$ of the total current in the circuit. If the resistance of the galvanometer is $G$, the shunt resistance $S$ connected across the galvanometer is:

$A$ student connects a voltmeter,an ammeter,and a resistor as shown in the circuit. If the voltmeter reading is $20 \ V$ and the ammeter reading is $4 \ A$,then the resistance $R$ will be:

$A$ galvanometer with a resistance of $8 \, \Omega$ is connected to a shunt of $2 \, \Omega$. If the total current is $1 \, A$,how much current in $A$ passes through the shunt?

In a moving coil galvanometer,to make the field radial,what is done?

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