Question

An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J. (a) If the electric motor draws 8.80 kW as the car moves at a steady speed of 20.0 m/s, what is the current (in A) delivered to the motor? A (b) How far (in km) can the car travel before it is "out of juice"? km (c) What If? The headlights of the car each have a 55.0 W halogen bulb. If the car is driven with both headlights on, how much less will its range be (in m)? m

Answer 1

Answer:

a) $I=733.33\ A$

b) $d=52272.7273\ m$

c) $d'=51948.0519\ m$

Explanation:

Given:

• voltage of the battery, $V=12\ V$

• energy storage capacity of the battery, $E=2.3\times 10^7\ J$

• speed of the car, $v=20\ m.s^{-1}$

a)

power drawn by the car, $P=8.8\ kW$

Now the Current delivered to the motor:

we the relation between the power and electrical current,

$P=V.I$

$8800=12\times I$

$I=733.33\ A$

b)

Distance travelled before battery is out of juice:

we first find the time before the battery runs out,

$t=\frac{E}{P}$

$t=\frac{2.3\times 10^7}{8800}$

$t=2613.636\ s$

Now the distance:

$d=v.t$

$d=20\times 2613.636$

$d=52272.7273\ m$

c)

When the head light of 55 W power is kept on while moving then the power   consumption of the car is:

$P'=P+55$

$P'=8800+55$

$P'=8855\ W$

Now the time of operation of the car:

$t'=\frac{E}{P'}$

$t'=\frac{2.3\times 10^7}{8855}$

$t'=2597.4026\ s$

Now the distance travelled:

$d'=v.t'$

$d'=20\times 2597.4025$

$d'=51948.0519\ m$