A galvanometer having a coil resistance 100 ; | vedclass

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(A) Given: Galvanometer resistance $R_{g} = 100 \; \Omega$,full-scale deflection current $i_{g} = 1 \; mA = 1 	imes 10^{-3} \; A$,and target potentia

Vedclass · Accepted Answer

$9.9$

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(A) Given: Galvanometer resistance $R_{g} = 100 \; \Omega$,full-scale deflection current $i_{g} = 1 \; mA = 1 	imes 10^{-3} \; A$,and target potential difference $V = 10 \; V$.To convert a galvanometer into a voltmeter,a high resistance $R$ is connected in series with the galvanometer.The formula for the potential difference is $V = i_{g}(R + R_{g})$.Substituting the given values:$10 = 1 	imes 10^{-3} 	imes (R + 100)$$10 / (1 	imes 10^{-3}) = R + 100$$10000 = R + 100$$R = 10000 - 100 = 9900 \; \Omega$.Converting to $k\Omega$: $R = 9900 / 1000 = 9.9 \; k\Omega$.

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