Current sensitivities of two galvanometers $\mathrm{G}_1$ and $\mathrm{G}_2$ of resistances $100 \Omega$ and…
- Same in both galvanometers
- More in $\mathrm{G}_2$
- Zero
- More in $\mathrm{G}_1$
Solution
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)