First,a set of n equal resistors of 10 ; Omeg | vedclass

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(A) In the series connection,the equivalent resistance of the resistors is $R_{eq,s} = nR = 10n \; \Omega$.The total resistance of the circuit is $R_{

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$20$

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(A) In the series connection,the equivalent resistance of the resistors is $R_{eq,s} = nR = 10n \; \Omega$.The total resistance of the circuit is $R_{total,s} = 10 + 10n \; \Omega$.The current in the series circuit is $I_s = \frac{20}{10 + 10n} = \frac{2}{1 + n} \; A$.In the parallel connection,the equivalent resistance of the resistors is $R_{eq,p} = \frac{R}{n} = \frac{10}{n} \; \Omega$.The total resistance of the circuit is $R_{total,p} = 10 + \frac{10}{n} = \frac{10n + 10}{n} \; \Omega$.The current in the parallel circuit is $I_p = \frac{20}{\frac{10n + 10}{n}} = \frac{20n}{10(n + 1)} = \frac{2n}{n + 1} \; A$.Given that the current increases $20$ times,we have $I_p = 20 I_s$.Substituting the expressions: $\frac{2n}{n + 1} = 20 	imes \frac{2}{n + 1}$.Canceling the common terms,we get $2n = 40$,which implies $n = 20$.

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