You are given a number of 22 Ω resistors, each capable of dissipating only 2.1 W without being destroyed.

Question

You are given a number of 22 Ω resistors, each capable of dissipating only 2.1 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a 22 Ω resistance that is capable of dissipating at least 8.3 W?

Answer 1

 Each of the resistors will be able to handle a bit less than 2.1W. To get a total capacity of 8.4W We need more than 1 branch. First, we need resistors to be added to in series so that the power in each is reduced to 2.1W/8.4W = 1/4 . R_B= 22Ω +22Ω +22Ω+22Ω= 88 Ω. This reduces the power below 2.1 W in the series. 8.4W/2.1W =4. Then, in order to get the power we need, 4 branches must be put in parallel to carry only 1/4th the power. each . So we need 4 branches to match the power requirement.

1/R_T = (1/88Ω+1/88Ω+1/88Ω+1/88Ω. R_T=(88/4)Ω and RT= 22Ω.