Question

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

Answer 1

Answer:

• 0.73 kW

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:

$< power > = \frac{work}{time}$

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

$time = 45 \ min * 60 \frac{s}{min}$

$time = 2,700 \ s$

Taking all this together:

$< power > = \frac{1,980,000 \ J}{2,700 \ s}$

$< power > = \frac{1,980,000 \ J}{2,700 \ s}$

$< power > = 733.333 \ W$

or

$< power > = 0.733 \ kW$