Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

In order to pass 10% of the main current through a moving coil galvanometer of 99 ohms, the resistance of the required shunt is:

9.9 \Omega

10 \Omega

11 \Omega

9 \Omega

Principle: To convert a moving coil galvanometer into an ammeter, a low resistance called a 'shunt' ( $S$ ) is connected in parallel with the galvanometer coil. This arrangement allows the majority of the current to bypass the sensitive galvanometer coil .
Parallel Connection: Since the galvanometer ( $G$ ) and the shunt ( $S$ ) are in parallel, the potential difference across them is the same. $V_G = V_S \implies I_g G = (I - I_g) S$ Where: $I$ is the total main current. $I_g$ is the current passing through the galvanometer. $I - I_g$ is the current passing through the shunt. $G$ is the resistance of the galvanometer.
Given Data: Galvanometer Resistance $G = 99\ \Omega$ . Current through galvanometer $I_g = 10\%$ of $I = 0.10 I$ .

Calculation: Substituting the values into the equation: $(0.10 I) \times 99 = (0.90 I) \times S$ $S = \frac{0.10 \times 99}{0.90}$ $S = \frac{9.9}{0.9} = 11\ \Omega$ .
Conclusion: The required shunt resistance is $11\ \Omega$ .

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