Question

An electric car accelerates for 9.9 s by drawing energy from its 340-V battery pack. During this time, 1700 C of charge pass through the battery pack. Find the minimum power rating of the car

Answer 1

Step-by-step explanation:

Given:

Time, t = 9.9 s

Charge, Q = 1700 C

Potential difference, V = 340 V

Power = IV

I × t = Q

Power = (1700 × 3400)/9.9

= 58383.8 W

= 58.38 kW.

Answer 2

78.26hp

Step-by-step explanation:

q= 1700C

V=340V

t= 9.9s

1hp=746W

Energy obtained from the car= qv/t = 1700×340/9.9 × 1hp/746W= 78.26hp