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Question

Physics Question on Current electricity

A galvanometer has 30 divisions and a sensitivity 16μA/div16 \mu A/div . It can be converted into a voltmeter to read 3 V by connecting

A

resistance nearly 6kO6 kO in series

B

6KO6KO in parallel

C

500O500O in series

D

it cannot be converted

Answer

resistance nearly 6kO6 kO in series

Explanation

Solution

Current flowing through the galvanometer
Ig=16μA/div×30divI_g = 16 \mu A/div \times 30 div
=480μA= 480 \mu A
The value of resistance in series to the galvanometer is
R=VIgG=3480×1060R = \frac{V}{I_g} - G = \frac{3}{480 \times 10^{-6}} - 0
=6.25KO= 6.25KO