Question 241:
easyA galvanometer of $50\text{ }\Omega$ resistance has $25$ divisions. A current of $4 \times 10^{-4}\text{ A}$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25\text{ V}$, it should be connected with a resistance of: (2004)
Full scale current $I_g = 25 \times 4 \times 10^{-4} = 0.01\text{ A}$. Required series resistance $R = \frac{V}{I_g} - G = \frac{25}{0.01} - 50 = 2450\text{ }\Omega$.
Question 242:
easyTo convert a galvanometer into a voltmeter one should connect a:
(2002)
A voltmeter is formed by connecting a very high resistance in series with a galvanometer to measure large potential differences.
Question 243:
easyResistance of a galvanometer coil is $8\text{ }\Omega$ and $2\text{ }\Omega$ shunt resistance is connected with it. If main current is $1\text{ A}$ then the current flowing through $2\text{ }\Omega$ resistance will be:
(1998)
Current through shunt resistance $I_s = I \times \frac{G}{G + S} = 1 \times \frac{8}{8 + 2} = 0.8\text{ A}$.
Question 244:
easyTo convert a galvanometer into an ammeter, one needs to connect a:
(1992)
An ammeter is constructed by connecting a low resistance shunt in parallel with the galvanometer to bypass most of the current.