CK-12-Physics - Intermediate

17.4. Resistors in Series and Parallel http://www.ck12.org

Power=worktime→P=Wt Equation A.

Recall that the workWdone in moving a chargeqacross a potential differenceVis

W=qV.

SubstitutingqVfor work in Equation A, we haveP=Wt =qVt =qtV=IV→Pelectric power=IVRecall that charge

per unit time is currentqt=Ithe current.

Thus,

1 W= 1 A∗V or 1 C∗sV

In the case of the “plug” mentioned earlier, the power output is the result of a steady (not alternating) voltage of

7. 5 V, which produces a direct current of 200mA( 200 × 10 −^3 A).

The power is therefore

P=IV= ( 200 × 10 −^3 A)( 7. 5 V) = 1. 5 W.

Check Your Understanding

A 60-W incandescent light bulb is powered by a typical “120V AC” household electrical outlet. What average

current passes through the bulb?

Note: 120 VAC is an average value.

Answer:P=IV→ 60 W=I( 120 V)→I= 12060 WV= 0. 50 A

Illustrative Example 17.4.2

a. Utility companies charge customers in units of kilowatt-hours(kW h). Do utility companies charge for energy

consumption or power consumption?

Answer:

Unit analysis gives

kW h=energytimetime=energy

b. How much energy is 1. 0 kW h?

Answer:

1 KW= 1000 W=^1000 s J

1 H=60 min× 160 mins = 3 , 600 s→

1 kW h=^1000 s J× 3 , 600 s= 3. 6 × 106 J.

c. What is the cost for running a 1,500-W heater for 10 hours, if the cost perkW his 12 cents?

Answer:

1 , 500 W= 1. 5 kW

Energy used equals:E=kW h= 1. 5 kW× 10 H= 15 kW h→

15 kW h×$0kW h.^12 =$1. 80

During a cold winter month, if this heater is used for 10 hours per day for an entire month, the cost is $54.00 for the

month, just for this one device!