For designing a voltmeter of range 50,V and a | vedclass

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(C) For the voltmeter:The resistance $R$ to be connected in series is given by $R = \frac{V}{I_g} - G$.Given $V = 50\,V$,$I_g = 1\,mA = 10^{-3}\,A$,an

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$(A)$ and $(C)$

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(C) For the voltmeter:The resistance $R$ to be connected in series is given by $R = \frac{V}{I_g} - G$.Given $V = 50\,V$,$I_g = 1\,mA = 10^{-3}\,A$,and $G = 54\,\Omega$.$R = \frac{50}{10^{-3}} - 54 = 50000 - 54 = 49946\,\Omega \approx 50\,k\Omega$.Thus,statement $(A)$ is correct.For the ammeter:The shunt resistance $r$ to be connected in parallel is given by $r = \frac{I_g G}{I - I_g}$.Given $I = 10\,mA = 10^{-2}\,A$,$I_g = 1\,mA = 10^{-3}\,A$,and $G = 54\,\Omega$.$r = \frac{10^{-3} 	imes 54}{10 	imes 10^{-3} - 1 	imes 10^{-3}} = \frac{54 	imes 10^{-3}}{9 	imes 10^{-3}} = 6\,\Omega$.Thus,statement $(C)$ is correct.Therefore,the correct option is $(A)$ and $(C)$.

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