To reduce the current flowing through a coil | vedclass

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(D) Let the initial current be $I$ and the resistance of the coil be $G = 90\, \Omega$.We want the current through the coil $(I_g)$ to be reduced by $

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$10$

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(D) Let the initial current be $I$ and the resistance of the coil be $G = 90\, \Omega$.We want the current through the coil $(I_g)$ to be reduced by $90\%$,which means the new current $I_g = I - 0.9I = 0.1I = I/10$.Using the formula for shunt resistance $S$ required to divert current such that the galvanometer current becomes $I_g = I/n$,where $n = I/I_g = 10$.The shunt resistance is given by $S = \frac{G}{n - 1}$.Substituting the values: $S = \frac{90}{10 - 1} = \frac{90}{9} = 10\, \Omega$.Therefore,a resistor of $10\, \Omega$ must be connected in parallel.

To reduce the current flowing through a coil of resistance $90\, \Omega$ by $90\%$,a resistor of what value in $\Omega$ must be connected in parallel?

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