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:

Option 1 - 1000

Option 2 - 10

Option 3 - 100

Option 4 - 1

3 Views|Posted 5 months ago

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Answered by

Raj Pandey | Contributor-Level 9

5 months ago

Correct Option - 3

Detailed Solution:

$i_{series} = \frac{E}{R_{series}}$

$\Rightarrow i_{series} = \frac{E}{10 R}$

$i_{parallel} = \frac{E}{R_{Parallel}}$

$= \frac{E}{R / 10}$

$i_{parallel} = n * i_{series}$

$\Rightarrow \frac{10 E}{R} = \frac{n E}{10 R}$

$\Rightarrow n = 100$

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Answered by

Raj Pandey | Contributor-Level 9

5 months ago

$i_{series} = \frac{E}{R_{series}}$

$? i_{series} = \frac{E}{10 R}$

$i_{parallel} = \frac{E}{R_{Parallel}}$

$= \frac{E}{R / 10}$

$i_{parallel} = n * i_{series}$

$? \frac{10 E}{R} = \frac{n E}{10 R}$

$? n = 100$

Upvote

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Physics Current Electricity 2025

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