91影视

Ohm鈥檚 law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

V =IR

According to Kirchhoff鈥檚 Lawthe sum of the voltages around the closed loop is equal to null.

$\underset{\mathbf{}}{\mathbf{鈭�}}\mathit{V}\mathbf{=}\mathbf{0}$

Resistance of a wire is given by

$\mathit{R}\mathbf{=}\mathit{蚁}\frac{\mathbf{L}}{\mathbf{A}}$

where $\mathit{蚁}$ is the resistivity and Lis the length of the wire and A is the area of the wire

By use of ohm鈥檚 law we have$I=\frac{V}{R}$

Replace R by$R=蚁\frac{L}{A}$

Get

$I=\frac{VA}{pL}$

Since the voltage, length and the resistivity remain the same we get

$I=f\left(A\right)=f\left(D\right)$

Where D is the diameter of the wire.

Hence, We consider the diameter of the wire

We know that the power is given by

P=VI

Rearranging we get

$I=\frac{P}{V}$

We know that $\mathrm{P}=4200\mathrm{W}\mathrm{and}\mathrm{V}=120\mathrm{V}$

Input this

$I=\frac{4200W}{120V}=35A$

Look up the table and conclude that Gauge 8 is suitable to carry this load since it can have maximum current of 40A.

We know that the power dissipated is given by

$P=I^{2}R$

We know $R=蚁\frac{L}{A}$, substitute this to get

$$\begin{gathered} P=I^2蚁\frac{L}{A} \\ P=I^2蚁\frac{L}{蟺r_8^2} \end{gathered}$$

Use $蚁=1.72脳{10}^{-8}惟m$and $r_8$is the radius of gauge 8 wire and length is 42m

Input this data into the formula we get

P =106 W

Hence, The power dissipated is P=106 W

Use the radius for gauge 6 wire to find power consumption of a gauge 6 wire.

$P=I^{2}p\frac{L}{{\mathrm{蟺谤}}_6^2}$

The saving on an appliance would be

$$\begin{gathered} P_s=P_8-P_6 \\ {P}_s=106W-66W \\ {P}_s=40W \end{gathered}$$

Since the appliance is kept on for 12 hours per day multiply it by 365 days to obtain number of hours it works

$t=\raisebox{1ex}{$4380h$}\!\left/ \!\raisebox{-1ex}{$year$}\right.$

Multiply this by the power saving and cost of one Kwh

$$\begin{gathered} \raisebox{1ex}{$SavingCost=\left(40W\right)(\$0.11)(4380h$}\!\left/ \!\raisebox{-1ex}{$year$}\right. \\ SavingCost=\$19.25 \end{gathered}$$

Average Saving cost is $19.25