NEET 2023 — Physics solution

NEET2023Physics

$10$ 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:

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

Solution

Series: $R_{s} = 10 R$ , so $I_{s} = \frac{E}{10 R}$ .

Parallel: $R_{p} = \frac{R}{10}$ , so $I_{p} = \frac{E}{R /10} = \frac{10 E}{R}$ .

$n = \frac{I _{p}}{I _{s}} = \frac{10 E / R}{E / ( 10 R )} = 100$ .

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