Question

An electric car accelerates for 8.0 s by drawing energy from its 320-V battery pack. During this time, 1300 C worth of electrons pass through the battery pack. (a) How many electrons are moved through the battery during this 8.0 s acceleration time? (b) How much energy transfer does this constitute? (c) Find the minimum horsepower rating of the car.(746 W = 1 hp)

Answer 1

(a)
`8.13\cdot 10^(-21)`

The magnitude of the charge of one electron is

`q=1.6\cdot 10^(-19)C`

Here the total amount of charge that passed through the battery pack is

Q = 1300 C

So this total charge is given by

Q = Nq

where

N is the number of electrons that has moved through the battery

Solving for N,

`N=(Q)/(q)=(1300 C)/(1.6\cdot 10^(-19) C)=8.13\cdot 10^(-21)`

(b)
`4.16\cdot 10^5 J`

First, we can find the current through the battery, which is given by the ratio between the total charge (Q = 1300 C) and the time interval (t = 8.0 s):

`I=(Q)/(t)=(1300 C)/(8.0 s)=162.5 A`

Now we can find the power, which is given by:

`P=VI`

where

V = 320 V is the voltage

I = 162.5 A is the current

Subsituting,

`P=(320 V)(162.5 A)=52,000 W`

And now we can find the total energy transferred, which is the product between the power and the time:

`E=Pt = (52,000 W)(8.0 s)=4.16\cdot 10^5 J`

(c) 69.7 hp

Now we have to convert the power from Watt to horsepower.

We know that

1 hp = 746 W

So we can set up the following proportion:

1 hp : 746 W = x : 52,000 W

And by solving for x, we find the power in horsepower:

`x=(1 hp \cdot 52,000 W)/(746 W)=69.7 hp`