Question: You are given a number of10$\mathit{\Omega }$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 a10resistance that is capable of dissipating at least 5.0 W?

The equivalent resistor is $R_{eq}=10\Omega$.

You can write for the equivalent resistance of n resistors in series. Then from the given condition, you can find the number of branches of series complexes that are connected in parallel. Then from the rate of dissipation of a single resistor, you can find the minimum number of resistances that we need to connect in series and parallel to make a 10 $\Omega$ resistance that is capable of dissipating at least 5.0 w .

Formulae:

Equivalent resistance when n resistors R are in series:

$R_{eq}=nR$

Equivalent resistance when resistors are in parallel is,

$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}$

There are n resistance in series, so their equivalent resistance is,

$R_{eq}=nR$

Let’s assume that there are m numbers of such branches present in that circuit.

So, the equivalent resistance is

$$\begin{gathered} \frac{1}{R_{eq}}=\sum _{i=1}^m\frac{1}{nR}+\frac{1}{nR}+\dots . \\ =\frac{m}{nR} \end{gathered}$$

Since, the condition given in the problem is,

$R_{eq}=R$

Therefore,

This implies that the number of resistors in series is equal to the number of branches of the series complexes that are connected in parallel.

m = n

Suppose there are 2 resistors in series and 2 branches of such 2 resistors in series connected in parallel. That means there are 4 total resistors in the whole circuit.

And the problem says that each resistor dissipates 1.0W power, so such a circuit dissipates 4.0 W power.

But if we connect 3 resistors in series and 3 such branches, then there will be 9 resistors, and that will dissipate 9.0 W power. That is more than 5.0 W .

n = 9

So,the minimum number of resistance is,

n = 9