Question

Three 40-Ω lightbulbs and three 80-Ω lightbulbs are connected in series.(a) What is the total resistance?(b) What would be their resistance if all six were wired in parallel?

Answer 1

(a)

In order to find the series total resistance, we need to add all resistances:

$R_{eq}=R_1+R_2+R_3+R_4+R_5+R_6$

If three resistances are 40 ohms and three are 80 ohms, we have:

$\begin{gathered} R_{eq}=40+40+40+80+80+80\\ \\ R_{eq}=360\text{ ohms} \end{gathered}$

(b)

Now, since the resistances are in parallel, we need to use the expression below:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}+\frac{1}{R_6}$

So we have:

$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{40}+\frac{1}{40}+\frac{1}{40}+\frac{1}{80}+\frac{1}{80}+\frac{1}{80}\\ \\ \frac{1}{R_{eq}}=\frac{3}{40}+\frac{3}{80}\\ \\ \frac{1}{R_{eq}}=\frac{6}{80}+\frac{3}{80}\\ \\ \frac{1}{R_{eq}}=\frac{9}{80}\\ \\ R_{eq}=\frac{80}{9}\text{ ohms} \end{gathered}$