Question

A lightweight electric car is powered by a series combination of ten 12.0-V batteries, each having negligible internal resistance. Each battery can deliver a charge of 154 A · h before needing to be recharged. At a speed of 82 km/h, the average frictional force due to air drag and rolling friction is 1.34 kN.

(a) What must be the minimum power delivered by the electric motor if the car is to travel at a speed of 82 km/h?
________kW

(b) What is the total charge, in coulombs, that can be delivered by the series combination of ten batteries before recharging is required?
_________kC

(c) What is the total electrical energy delivered by the 10 batteries before recharging?
______MJ

(d) How far can the car travel (at 82 km/h) before the batteries must be recharged?
________ km

(e) What is the cost per kilometer if the cost of recharging the batteries is 9 cents per kilowatt-hour?

_____$/km

Answer 1

▪︎Givens :

Voltage, V = 12 V

Charge capacity,   = 154 Ah

velocity, v = 82 km/h

Frictional force, F = 1.34 kN

a) Power = Force × velocity

= 1.34 kN × 82 km/h

      = 1340 N × (82000m/3600s)

= 30522.2 Nm/s = 30522.2 J/s = 30522.2 W

  = 30.5 kW

b) Batteries in series do not increase the capacity.

(If you wired up two 12V batteries each with acapacity of 20Ah, the total energy is 12 × 20 × 2 = 480 VAh. If you wired up one 24V battery of capacity 20Ah, the total energy is 24 ×20 = 480 VAh. However, if you wired up 24V battery with capacity 40Ah, total energy is 960VAh).

Total charge = 154 Ah = 154 C/s × 3600s = 554400 C

= 554.4 kC

c) Total electrical charge,U = voltage × charge

U = (12 × 10 ) × 554400

= 66.5 × 10⁶ J

= 66.5 MJ

d) Distance = speed × time

Time = energy ÷ power

= 2179.66 seconds ÷ 3600

= 0.6 hours

Distance = 82 km/h × 0.6h = 49.6 km

e) 1kWh = 1000 J/s × 3600s = 3.6 × 10⁶ J

0.09$ ÷ 1kWh = 0.09$ ÷ ( 3.6 × 10⁶ ) J

= 2.5 × 10^-8 $/J

49.6 km ÷ (66.5 × 10⁶) J = 7.46 × 10^-7 km/J

$/km = ($/J) × (J/km)

= 2.5 × 10^-8 ÷ 7.46 × 10^-7

   = 0.03 $/km