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2 Answers

Saurabh Kumar
askIITians Faculty 2411 Points
6 years ago
Let the resistors be divided into groups of n resistors each, with all the resistors in the same group connected in series. Suppose there are m such groups that are connected in parallel with each other. Let R be the resistance of any one of the resistors. Then the equivalent resistance of any group is nR, and Req, the equivalent resistance of the whole array, satisfies,
1/Reqv = m/nR
Since the problem requires Req = 10 ohm = R, we must select n = m. We note that the current is the same in every resistor and there are n x m = n2 resistors, so the maximum total power that can be dissipated is Ptotal = n2P, where is the maximum power that can be dissipated by any one of the resistors. The problem demands Ptotal ³ 5.0P, so n2 must be at least as large as 5.0. Since n must be an integer, the smallest it can be is 3. The least number of resistors is n2 = 9.

35 Points
6 years ago
i didn’t catch your point..can you please explain this....

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