Current sensitivities of two galvanometers $\mathrm{G}_1$ and $\mathrm{G}_2$ of resistances $100 \Omega$ and…

Current sensitivities of two galvanometers $\mathrm{G}_1$ and $\mathrm{G}_2$ of resistances $100 \Omega$ and $50 \Omega$ are $10^8 \mathrm{div} / \mathrm{A}$ and $0.5 \times 10^5 \mathrm{div} / \mathrm{A}$ respectively. The galvanometer in which the voltage sensitivity is more is
  1. Same in both galvanometers
  2. More in $\mathrm{G}_2$
  3. Zero
  4. More in $\mathrm{G}_1$

Solution

$\mathrm{G}_1=100 \Omega, \mathrm{G}_2=50 \Omega$ $\mathrm{I}_{\mathrm{S} 1}=10^8 \mathrm{div} / \mathrm{A}, \mathrm{I}_{\mathrm{S} 2}=0.5 \times 10^5 \mathrm{div} / \mathrm{A}$
Voltage sensitivity, $V_S=\frac{I_S}{G}$ $\begin{aligned} & \therefore \mathrm{V}_{\mathrm{S} 1}=\frac{\mathrm{I}_{\mathrm{S} 1}}{\mathrm{G}_1}=\frac{10^8}{100}=10^6 \mathrm{~V} / \mathrm{div} \\ & \mathrm{~V}_{\mathrm{S} 2}=\frac{\mathrm{I}_{\mathrm{S} 2}}{\mathrm{G}_2}=\frac{0.5 \times 10^5}{50}=10^3 \mathrm{~V} / \mathrm{div} \\ & \therefore \quad \mathrm{~V}_{51}\gt\mathrm{V}_{52} \end{aligned}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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