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(B) In the half-deflection or fractional-deflection method,the galvanometer is connected in parallel with a shunt resistance $S$.Since the galvanomete

Vedclass · Accepted Answer

The $\frac{1}{3}$ deflection method can be used,and in this case,$G$ equals twice the value of the shunt resistance $(S)$.

Vedclass · Answer

(B) In the half-deflection or fractional-deflection method,the galvanometer is connected in parallel with a shunt resistance $S$.Since the galvanometer and shunt are in parallel,the potential difference across them is the same:$(I - I_g) S = I_g G$Where $I$ is the total current,$I_g$ is the current through the galvanometer,and $G$ is the galvanometer resistance.Rearranging the terms,we get the ratio of currents as:$\frac{I_g}{I} = \frac{S}{S + G}$Given that the deflection is $\frac{1}{3}$ of the initial value,the current through the galvanometer becomes $I_g = \frac{1}{3} I$,which implies $\frac{I_g}{I} = \frac{1}{3}$.Substituting this into the equation:$\frac{1}{3} = \frac{S}{S + G}$$S + G = 3S$$G = 2S$Thus,the resistance of the galvanometer is twice the value of the shunt resistance used.

$A$ teacher in his physics laboratory allotted an experiment to determine the resistance $(G)$ of a galvanometer. Students took the observations for $\frac{1}{3}$ deflection in the galvanometer. Which of the following is true for measuring the value of $G$?

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