JEE MAIN - Physics (2021 - 27th August Morning Shift - No. 22)
First, a set of n equal resistors of 10 $$\Omega$$ each are connected in series to a battery of emf 20V and internal resistance 10$$\Omega$$. A current I is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of n is ............... .
Answer
20
Explanation
In series
$${R_{eq}} = nR = 10n$$
$${i_s} = {{20} \over {10 + 10n}} = {2 \over {1 + n}}$$
In parallel
$${R_{eq}} = {{10} \over n}$$
$${i_p} = {{20} \over {{{10} \over n} + 10}} = {{2n} \over {1 + n}}$$
$${{{i_p}} \over {{i_s}}} = 20$$
$${{\left( {{{2n} \over {1 + n}}} \right)} \over {\left( {{2 \over {1 + n}}} \right)}} = 20$$
$$n = 20$$
$${R_{eq}} = nR = 10n$$
$${i_s} = {{20} \over {10 + 10n}} = {2 \over {1 + n}}$$
In parallel
$${R_{eq}} = {{10} \over n}$$
$${i_p} = {{20} \over {{{10} \over n} + 10}} = {{2n} \over {1 + n}}$$
$${{{i_p}} \over {{i_s}}} = 20$$
$${{\left( {{{2n} \over {1 + n}}} \right)} \over {\left( {{2 \over {1 + n}}} \right)}} = 20$$
$$n = 20$$
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