Question

You have a $2.0\Omega$resistor, a $3.0\Omega$resistor, a $6.0\Omega$resistor, and a $6.0\mathrm{V}$battery. Draw a diagram of a circuit in which all three resistors are used and the battery delivers 9.0 W of power.

Short answer

Diagram of the circuit

Step-by-step solution

Given information

We have given

Resistances =$2\Omega ,3\Omega ,6\Omega$

Battery Voltage =$6\mathrm{V}$

Power =$9\mathrm{W}$

We have to draw the correct circuit diagram.

Simplify

we know that power delivered by the battery will be,

$$\begin{gathered} \mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}} \\ \mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}}=\frac{36}{9}=4\Omega \end{gathered}$$

we should arrange the resistance just like that we get total equivalent is localid="1649324068838" $4\Omega .$

There is only one method if we arrange the two of them in parallel and one in series then,

localid="1649324073026" $$\begin{gathered} \frac{1}{{\mathrm{R}}_{\mathrm{eq}1}}=\frac{1}{3}+\frac{1}{6} \\ {R}_{eq1}=2\Omega \\ {R}_{eq}=2+2=4\Omega \end{gathered}$$