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 dissipated in the n resistors in both circuit is same. The ratio $R_{0} / R$ is :

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n^{2}

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1/n

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Solution

The correct option is A 1
Answer is A.
The current in the ciruits is given as $i_{1} = \frac{n E}{n R 0 + R} a n d i_{2} = \frac{E}{R 0 + n R}$ .
The power in the circuit is given as $P_{1} = \frac{n E^{2} R}{{(n R 0 + R)}^{2}} a n d P_{2} = \frac{n E^{2} R}{{(R 0 + n R)}^{2}}$ .
Since, $P_{1} = P_{2}$
Hence, the ratio, R0/R = 1.

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The correct option is A 1Answer is A.The current in the ciruits is given as i1=nEnR0+Randi2=ER0+nR.The power in the circuit is given as P1=nE2R(nR0+R)2andP2=nE2R(R0+nR)2.Since, P1=P2Hence, the ratio, R0/R = 1.