The current flowing in a coil of resistance 9 | vedclass

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(D) Let the initial current in the coil be $I$. We want to reduce the current by $90\%$,so the new current $I'$ flowing through the coil will be $I' =

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$10$

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(D) Let the initial current in the coil be $I$. We want to reduce the current by $90\%$,so the new current $I'$ flowing through the coil will be $I' = I - 0.9I = 0.1I = \frac{I}{10}$. Let $G = 90 \,\Omega$ be the resistance of the coil and $S$ be the shunt resistance connected in parallel. The current $I'$ through the coil is given by the current divider rule: $I' = I \left( \frac{S}{G + S} \right)$. Substituting the values: $\frac{I}{10} = I \left( \frac{S}{90 + S} \right)$. Canceling $I$ from both sides: $\frac{1}{10} = \frac{S}{90 + S}$. Cross-multiplying gives: $90 + S = 10S$. $9S = 90$,which implies $S = 10 \,\Omega$.

The current flowing in a coil of resistance $90 \,\Omega$ is to be reduced by $90\%$. What value of resistance should be connected in parallel with it? ............. $\Omega$

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