Problem 3: A current of 6A is drawn from a 120V line. What power is being developed? How much energy in joule and kWh is expended if the current is drawn steadily for one week?

Solution:

Step: 1 Overview

Power dissipation in resistors:
Power is a rate of doing work. It can also be defined as the ratio of work done and time.

P=Work/time
P=W/t
P=QV/t (because Work=QV)
P=V(Q/t)
P=VI

Energy in kWh:
Energy in commercial purpose measured in kWh, which is a very large unit as compared to Joule. Our electric measuring meters show reading in this unit. To calculate energy in kWh we have to convert time into hours, and power into kilo watts.

Step: 2 Calculation

Given:

I=6A
V=120V
time (t) = 1wk = (7x24)h = 168h = (168x3600)s = 604800s

Required:

P=?
E=?

Solution:

We know that:
P=VI
P=(120)(6)
P=720 Watts (Ans)

Also We know that E=Pt
E=(720W)(604800s)
E=4.35 x 10⁸J (Ans)

Again E=Pt
E=(720W)(168h)
E=(0.72kW)(168h)
E=120.96kWh (Ans)