Question

Low voltage power system: You want to "supercharge"1 your brand new Tesla Model 3 at a rate of 100 kW at home. For safety reasons, you chose a 5 V direct current (DC) voltage source. a) Calculate the required current. b) Assuming a cylindrical conductor with a current density of 10 A/mm2 , what diameter conductor would you need? c) Assuming copper with a resistivity of rhoCu = 1.67×10−8 Ωm and a cable length

Answer 1

(a) 20 kA (B)
`diameter=5.0035* 10^(-2)m` (c) length = 29.94 m

Step-by-step explanation:

We have given power P = 100 KW

Voltage V = 5 V

Current density
`10A/m^2=10* 10^(6)A/m^2`

(A) We have to calculate current

We know that power P = voltage × current

So
`100* 1000=5* i`

`i=20kA`

(b) We know that current density is given by

Current density
`=(current)/(area)`

So
`10* 10^(6)A/m^2=(20* 10^3)/(area)`

`area=2* 10^(-3)m^2`

We know that
`a=\pi r^2`

`2* 10^(-3)=3.14* r^2`

`r^2=6.34* 10^(-4)`

`r=2.517* 10^(-2)m`

`d=2 r=2* 2.517* 10^(-2)=5.035* 10^(-2)m`

(c) We have to find the length of the cable

We know that resistance
`R=(V)/(I)=(5)/(20* 10^3)=0.25mohm`

Resistance is given by
`R=\rho (l)/(A)`

So
`0.25* 10^(-3)=1.67* 10^(-8)(l)/(2* 10^(-3))`

l = 29.94 m