(C) In the series combination,the total resistance is $R_S = 10R$. The current $I_S$ is given by $I_S = \frac{E}{10R}$.In the parallel combination,the

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(C) In the series combination,the total resistance is $R_S = 10R$. The current $I_S$ is given by $I_S = \frac{E}{10R}$.In the parallel combination,the equivalent resistance is $R_P = \frac{R}{10}$. The current $I_P$ is given by $I_P = \frac{E}{R/10} = \frac{10E}{R}$.According to the problem,the current increases $n$ times,so $I_P = n \times I_S$.Substituting the values: $\frac{10E}{R} = n \times \frac{E}{10R}$.Solving for $n$: $n = \frac{10E}{R} \times \frac{10R}{E} = 100$.Therefore,the value of $n$ is $100$.

Class 12 Physics Current Electricity Equivalent Resistance - Series and Parallel , Circuit

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$10$ resistors,each of resistance $R$,are connected in series to a battery of $emf$ $E$ and negligible internal resistance. When those are connected in parallel to the same battery,the current is increased $n$ times. The value of $n$ is:

NEET 2023 All NEET

A
$1000$
B
$10$
C
$100$
D
$1$

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Current Electricity

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Equivalent Resistance - Series and Parallel , Circuit

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$10$ resistors,each of resistance $R$,are connected in series to a battery of $emf$ $E$ and negligible internal resistance. When those are connected in parallel to the same battery,the current is increased $n$ times. The value of $n$ is:

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