Question

It is necessary to deliver a current of 8 A in a circuit whose resistance is 5Ω. What is the least number of accumulators that must be used for this purpose and how should they be connected if the e.m.f. of each accumulator is 2V and the internal resistance is 0.5Ω.

Answer 1

Answer: TV = 160 accumulators. The battery should consist of 40 series-connected groups, with four accumulators connected in parallel in each group; or 4 parallel-connected groups with 40 series-connected accumulators in each group.

Explanation:

The current delivered by the battery is

A battery consisting of N cells delivers the largest current when its internal resistance is equal to the resistance of the external circuit (see the solution of Example 397), i.e., when in the denominator of equation (1) we have

The values found for N and m thus determine the least number of accumulators which must be used, and the manner in which they are connected.