Question

Suppose you have resistors $2.5\mathrm{k}\Omega$, $3.5\mathrm{k}\Omega$and $4.5\mathrm{k}\Omega$ and a $100\mathrm{V}$ power supply. What is the ratio of the total power delivered to the resistors if they are connected in parallel to the total power delivered if they are connected in series?

Short answer

The ration will be$14:5$

Step-by-step solution

Given information

We have given,

Voltage = $100\mathrm{V}$

Resistances =$2.5\mathrm{k}\Omega ,3.5\mathrm{k}\Omega ,4.5\mathrm{k}\Omega$

We have to find the ratio of power dissipated by when resistances are series and when are in parallel.

Simplify

When resistances are in parallel,

$$\begin{gathered} \frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{2.5}+\frac{1}{3.5}+\frac{1}{4.5} \\ {R}_{eq}=3.75k\Omega \end{gathered}$$

Then power is dissipated by the circuit,

localid="1649283398345" ${\mathrm{P}}_1=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{{\left(100\right)}^2}{3.75}$

When resistances are in series

${\mathrm{R}}_{\mathrm{eq}}=2.5+3.5+4.5=10.5\mathrm{k}\Omega$

Then power is dissipated by the circuit,

${\mathrm{P}}_2=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{{\left(100\right)}^2}{10.5}$

Divided the both power

ratio=$14:5$