Electrical Instruments

Let

G

be resistance of galvanometer and

i_g

the current through it. Let

V

is maximum potential difference, then from Ohm?? law

i_%7Bg%7D%20%3D%20%5Cfrac%7BV%7D%7BG%20%2B%20R%7D

%5CRightarrow%20%5C%2CR%20%3D%20%5Cfrac%7BV%7D%7Bi_%7Bg%7D%7D%20-%20G

Given,

G%20%3D%2010%5C%2C%5COmega%2C%20i_%7Bg%7D%20%3D%200.01%5C%2CA

V%20%3D%2010

volt

%5Ctherefore%20%5C%2C%20R%3D%5Cfrac%7B10%7D%7B0.01%7D-10%3D990%5C%2C%20%5COmega

Thus, on connecting a resistance

R

990%20%5COmega

in series with the galvanometer, the galvanometer will become a voltmeter of range zero to

10%5C%2C%20V

.
For the voltmeter, a high resistance is series with the galvanometer and so the resistance of a voltmeter is very high compared to that of galvanometer.

High-Yield Trend

Chapter Questions
1 MCQs

Official Solution

Electrical Instruments

High-Yield Trend

Chapter Questions 1 MCQs

Official Solution

Chapter Questions
1 MCQs