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An external circuit whose resistance is 0.3Ω is connected to six accumulators, each having an e.m.f. of 2V and an internal resistance of 0.2Ω. The accumulators are connected in similar groups in series and the groups are connected to each other in parallel. How should the accumulators be connected to obtain the largest current in the circuit? What will be the largest value of the current? 

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Answer: In each group there should be three series-connected accumulators. I = 10 amperes.

Explanation:

If each group contains m series-connected accumulators, then the resistance and e.m.f. of each group are mr0 and mε0 respectively. If the total number of accumulators, is N, then there will be N\m groups in the battery, and the resistance and e.m.f. of the battery, will be mr0/N and mε0, respectively. 

The current in the circuit is determined from Ohm's law:

The value of m which gives the highest value for I is that which makes the denominator of the fraction the smallest. Since the product of the terms of the denominator does not depend on m and is constant (mr0NR/m = NRr0), the denominator will have the smallest value when its terms are equal to each other i.e. when 

This can be seen from the following argument. We want to show that an expression of the type ma + b/m, where a and b are positive constants, is a minimum when ma = b/m, i.e., when m2 = b/a. For this value of m the expression ma + b/m has the value l\/ab. In other words, we must show that the inequality ma + b/m ≥ 2ab holds for all positive values of m. Squaring both sides, we obtain

The left-hand side of the inequality is equal to (ma — b/m)2, which is always a positive quantity. Therefore the inequality holds for all positive values of m, which is what we wished to prove.

or, in other words, when the internal resistance of the battery is equal to the resistance of the external circuit. Therefore, the denominator will be smallest, and, consequently, the current will be the largest, when

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