10 resistors, each of resistance are connected in series to a battery of emf and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased times. The value of is [2023]

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Solution

Q.
10 resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased $n$ times. The value of $n$ is [2023]

1 1

2 1000

3 10

4 100

Ans.
(4)

Net resistance in series combination,

$R_{N} = R_{1} + R_{2} + \dots + R_{10}$

$= 10 + 10 + \dots + 10 = 10 \times 10 = 100 Ω$ ...(i)

Same number of resistors and each of same magnitude connected in parallel combination,

$R'_{N} = {(\frac{1}{R_{1}} + \frac{1}{R_{2}} + \dots + \frac{1}{R_{10}})}^{- 1}$

$= {(\frac{1 + 1 + \dots + 1}{10})}^{- 1} = \frac{10}{10} = 1 Ω$ ...(ii)

Using (i) and (ii), $\frac{I'}{I} = \frac{V . R_{N}}{R'_{N} . V} = \frac{100}{1}$

Thus, option (4) is correct.

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