Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is:

40 \Omega

25 \Omega

250 \Omega

500 \Omega

Step-by-Step Solution

Definitions:

Current Sensitivity ( $I_s$ ) is the deflection per unit current: $I_s = \frac{\phi}{I}$ .
Voltage Sensitivity ( $V_s$ ) is the deflection per unit voltage: $V_s = \frac{\phi}{V}$ .

Relationship: According to Ohm's Law, $V = IR$ , where $R$ is the resistance of the galvanometer. Substituting $V$ into the voltage sensitivity equation: $V_s = \frac{\phi}{IR} = \frac{1}{R} \left( \frac{\phi}{I} \right) = \frac{I_s}{R}$ Rearranging for Resistance ( $R$ ): $R = \frac{I_s}{V_s}$
Calculation:

Given $I_s = 5 \text{ div/mA} = 5 \times 10^3 \text{ div/A}$ (converting mA to A).
Given $V_s = 20 \text{ div/V}$ .
Substitute the values: $R = \frac{5 \text{ div/mA}}{20 \text{ div/V}} = \frac{5}{20} \frac{\text{V}}{\text{mA}} = 0.25 \text{ k}\Omega = 250 \, \Omega$ Alternatively, using base units: $R = \frac{5000}{20} = 250 \, \Omega$ .

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