To determine the resistance G of a galvanomet | vedclass

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(A) In the first case,the current through the galvanometer is $I_g = \frac{V}{R + G}$.In the second case,when a shunt resistance $S$ is connected in p

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$S(R + G) = RG$

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(A) In the first case,the current through the galvanometer is $I_g = \frac{V}{R + G}$.In the second case,when a shunt resistance $S$ is connected in parallel with the galvanometer,the equivalent resistance of the parallel combination is $R_p = \frac{GS}{G + S}$.The total current in the circuit becomes $I = \frac{V}{R + R_p} = \frac{V}{R + \frac{GS}{G + S}}$.The current through the galvanometer is now $I_g' = I 	imes \frac{S}{G + S} = \frac{V}{R + \frac{GS}{G + S}} 	imes \frac{S}{G + S} = \frac{VS}{R(G + S) + GS}$.According to the problem,the deflection is halved,so $I_g' = \frac{I_g}{2}$.Substituting the expressions: $\frac{VS}{R(G + S) + GS} = \frac{1}{2} 	imes \frac{V}{R + G}$.$\frac{S}{RG + RS + GS} = \frac{1}{2(R + G)}$.$2S(R + G) = RG + RS + GS$.$2SR + 2SG = RG + RS + GS$.$2SR - RS + 2SG - GS = RG$.$SR + SG = RG$.$S(R + G) = RG$.

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