20.3k views
0 votes
What power in Kw is required to transfer a charge of 9000 coulomb through a potential difference of 220 volts in 45 min?

0.9 kW

0.73 kW

0.5 kW

1.6 kW

User Sureshraj
by
5.7k points

1 Answer

2 votes

Answer:

  • 0.73 kW

Step-by-step explanation:

The work W required for the transfer is equal to the energy difference


W = \Delta E = q \ \Delta V

where q is the charge we want to transfer and ΔV is the potential difference.

So, the work needed will be


W = 9,000 \ C * 220 V


W = 1,980,000 \ J

Now, the average power is:


< power > = (work)/(time)

As the Watt is Joule/second, we need the 45 min multiplied by 60:


time = 45 \ min * 60 (s)/(min)


time = 2,700 \ s

Taking all this together:


< power > = (1,980,000 \ J)/(2,700 \ s)


< power > = (1,980,000 \ J)/(2,700 \ s)


< power > = 733.333 \ W

or


< power > = 0.733 \ kW

User Milad Ahmadi
by
5.9k points