A galvanometer coil has a resistance of `12 Omega` and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

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Here `G=12 Omega, Ig=3mA=3xx10^(-3) A, V=18 V`
`R=(V)/(I_(g))-G=(18)/(3xx10^(-3))-12=(6000-12)Omega =5988 Omega.`

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