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 (integer)
20
Solution
In series<br><br>${R_{eq}} = nR = 10n$<br><br>${i_s} = {{20} \over {10 + 10n}} = {2 \over {1 + n}}$<br><br>In parallel<br><br>${R_{eq}} = {{10} \over n}$<br><br>${i_p} = {{20} \over {{{10} \over n} + 10}} = {{2n} \over {1 + n}}$<br><br>${{{i_p}} \over {{i_s}}} = 20$<br><br>$${{\left( {{{2n} \over {1 + n}}} \right)} \over {\left( {{2 \over {1 + n}}} \right)}} = 20$$<br><br>$n = 20$
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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