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:

`(1)/(R_(eq))=(1)/(R_1)+(1)/(R_2)+(1)/(R_3)+(1)/(R_4)+(1)/(R_5)+(1)/(R_6)`

So we have:

`\begin{gathered} (1)/(R_(eq))=(1)/(40)+(1)/(40)+(1)/(40)+(1)/(80)+(1)/(80)+(1)/(80)\\ \\ (1)/(R_(eq))=(3)/(40)+(3)/(80)\\ \\ (1)/(R_(eq))=(6)/(80)+(3)/(80)\\ \\ (1)/(R_(eq))=(9)/(80)\\ \\ R_(eq)=(80)/(9)\text{ ohms} \end{gathered}`