9. You are given a number of 10 Ω resistors, each capable of dissipating only 1.0 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 10 Ω resistance that is capable of dissipating at least 5.0 W?

here are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.

Explanation:

Let's work carefully this exercise, they indicate that the total resistance 10 ohm and dissipates 5W, so we can use the power equation to find the circuit current

            P = Vi = i² R

           i = √ P / R

            i = √ (5/10)

           i = 0.707 A

This is the current that must circulate in the circuit.

Let's build a circuit with three resistors in series and each resistor in series has three resistors in parallel

The equivalent resistance is

  1 / = 1/10 + 1/10 + 1/10 = 3/10

   Requi = 10/3

   Requi = 3.3 Ω

The current in the three series resistors is I = 0.707 A, and this is divided into three equal parts for the parallel resistors

current in each residence in parallel

              i_P = 0.707 / 3

              I_p = 0.2357 A

now let's look at the power dissipated in each resistor

            P = R i²

            P = 10  0.2357²

            P = 0.56 W

the power dissipated by each resistance is within the range of 1 A, let's see the total power that the 9 resistors dissipate

             P_total = 9 P

             P = total = 9 0.56

             P_total = 5 W

we see that this combination meets the specifications of the problem.

Therefore there are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more