A galvanometer of $40 \Omega$ resistance has a scale with 30 divisions. A current of $5 imes 10^{-

Question

A galvanometer of $40 \Omega$ resistance has a scale with 30 divisions. A current of $5 	imes 10^{-4} \mathrm{~A}$ gives a deflection of exactly one division. To convert this galvanometer into a voltmeter capable of reading up to $60 \mathrm{~V}$, it must be connected with a resistance of

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(b) :  The current required for full-scale deflection of the galvanometer is $I_g = 30 	imes 5 	imes 10^{-4} \mathrm{~A} = 150 	imes 10^{-4} \mathrm{~A} = 0.015 \mathrm{~A}$. To convert a galvanometer into a voltmeter of range $V$, a high resistance $R$ must be connected in series. The required resistance is $R = \frac{V}{I_g} - G = \frac{60}{0.015} - 40 = 4000 - 40 = 3960 \Omega$.