Current sensitivities of two galvanometers G_ | vedclass

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(D) The voltage sensitivity $(V_S)$ of a galvanometer is related to its current sensitivity $(I_S)$ and resistance $(R)$ by the formula: $V_S = \frac{

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(D) The voltage sensitivity $(V_S)$ of a galvanometer is related to its current sensitivity $(I_S)$ and resistance $(R)$ by the formula: $V_S = \frac{I_S}{R}$.For galvanometer $G_1$: Resistance $R_1 = 100 \ \Omega$,Current sensitivity $I_{S1} = 10^8 \ 	ext{div/A}$.$V_{S1} = \frac{10^8}{100} = 10^6 \ 	ext{div/V}$.For galvanometer $G_2$: Resistance $R_2 = 50 \ \Omega$,Current sensitivity $I_{S2} = 0.5 	imes 10^5 \ 	ext{div/A}$.$V_{S2} = \frac{0.5 	imes 10^5}{50} = \frac{50000}{50} = 1000 = 10^3 \ 	ext{div/V}$.Comparing the two,$V_{S1} = 10^6 \ 	ext{div/V}$ and $V_{S2} = 10^3 \ 	ext{div/V}$.Since $10^6 > 10^3$,the voltage sensitivity is higher in $G_1$.

Current sensitivities of two galvanometers $G_1$ and $G_2$ of resistances $100 \ \Omega$ and $50 \ \Omega$ are $10^8 \ \text{div/A}$ and $0.5 \times 10^5 \ \text{div/A}$ respectively. The galvanometer in which the voltage sensitivity is more is

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