In the circuit shown, n identical resistors R are connected in parallel (n > 1) and the combination in series to another resistor $R_{0}$ . In the adjoining circuit n resistors of resistance R are all connected in series along with $R_{0}$ .
The batteries in both circuits are identical and net power dissipate in the n resistors in both circuit is same. The ratio $R_{0} / R$ is

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n_{1}

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n^{2}

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1/n

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Solution

The correct option is A 1
In parallel combination
$i = (\frac{E}{R_{0} + \frac{R}{n}})$
$P_{1} = (I^{2} . \frac{R}{n}) = {(\frac{E}{R_{0} + \frac{R}{n}})}^{2} . \frac{R}{n}$
In series combination
$I = \frac{E}{R_{0} + n R}$
$P_{2} = {(\frac{E}{R_{0} + n R})}^{2} n R$
But $P_{1} = P_{2}$
${(\frac{E}{R_{0} + \frac{R}{n}})}^{2} . \frac{R}{n} = {(\frac{E}{R_{0} + n R})}^{2} . n R$
$\frac{n}{n R_{0} + R} = (\frac{n}{R_{0} + n R})$
$R_{0} + n R = n R_{0} + R$
$R_{0} (1 - n) = R (1 - n)$
$\frac{R_{0}}{R} = 1$

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The correct option is A 1In parallel combinationi=(ER0+Rn)P1=(I2.Rn)=(ER0+Rn)2.RnIn series combinationI=ER0+nRP2=(ER0+nR)2nRBut P1=P2(ER0+Rn)2.Rn=(ER0+nR)2.nRnnR0+R=(nR0+nR)R0+nR=nR0+RR0(1−n)=R(1−n)R0R=1