Question

You are given a number of 37 Ω resistors, each capable of dissipating only 3.8 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 37 Ω resistance that is capable of dissipating at least 11.3 W?

Answer 1

Let we have n resistors connected in series and m of these n resistors connected in parallel.

The equivalent resistance of n identical resistors in series:

Rs = nR

The equivalent resistance of m identical resistors connected in parallel with resistance of Rs resistors:

1/Req = 1/Rs + 1/Rs +.......+ 1/Rs (total m resistors)

=> Req = Rs/m = nR/m

But Req = R

Therefore n=m

Hence total number of resistors = n*m = n^2

Also Ptotal = n^2 Pr .......................(1)

Where Ptotal = 11.3 watt , Pr = 3.8 watt

=> Ptotal > or = 11.3 * Pr ....................(2)

Comparing (1) & (2),

n^2 > or = 11.3

but since n is the number of resistors which means n must be an integer

Therefore n^2 = 16

=> We need 16 resistors each one is capable of dissipating 3.8 watt to from a resistor capable of dissipating 11.3 watt.