A galvanometer of resistance 100,Omega has 50 | vedclass

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(D) The full-scale deflection current of the galvanometer is $I_g = 50 	imes 20 	imes 10^{-6} \, A = 10^{-3} \, A = 1 \, mA$.The resistance of the g

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(D) The full-scale deflection current of the galvanometer is $I_g = 50 	imes 20 	imes 10^{-6} \, A = 10^{-3} \, A = 1 \, mA$.The resistance of the galvanometer is $G = 100 \, \Omega$.For a voltmeter of range $V$,the total resistance required is $R_{total} = V / I_g$.The series resistance to be added is $R = R_{total} - G$.For $0-2 \, V$ range: $R_{total} = 2 / 10^{-3} = 2000 \, \Omega$. So,$R_1 = 2000 - 100 = 1900 \, \Omega$.For $0-10 \, V$ range: $R_{total} = 10 / 10^{-3} = 10000 \, \Omega$. The resistance added in series with $R_1$ is $R_2 = 10000 - 2000 = 8000 \, \Omega$.For $0-20 \, V$ range: $R_{total} = 20 / 10^{-3} = 20000 \, \Omega$. The resistance added in series with $R_1 + R_2$ is $R_3 = 20000 - 10000 = 10000 \, \Omega$.Thus,the circuit must have $R_1 = 1900 \, \Omega$,$R_2 = 8000 \, \Omega$,and $R_3 = 10000 \, \Omega$ in series.

$A$ galvanometer of resistance $100\,\Omega$ has $50$ divisions on its scale and a sensitivity of $20\,\mu A/\text{division}$. It is to be converted into a voltmeter with three ranges: $0-2\,V$,$0-10\,V$,and $0-20\,V$. The appropriate circuit to do so is:

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Assertion: Voltmeter is connected in parallel with the circuit.
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