1.

A maximum current of 0.5 mA can be passed through a galvanometer of resistance 20Omega. Calculate the resistance to be connected in series to convert it into a voltmeter of range (0- 5)V.

Answer»

Solution :R = G (n -1), where `n=V/(V_(g))`
`V=5V,V_(g)=i_(g)G=0.5xx10^(-3)xx20=10^(-2)V`
`thereforen=500` and `R=20(500-1)=9980Omega`


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