Solutions

 

1.         60 Hz = frequency

            120 V = potential difference

            1.7 A = current

            0.2 kW = power

 

2.         a.         3000 J

            b.         0.001 kJ

            c.         1.299 kJ

            d.         345 J

 

3.         a.         10 W

            b.         800 W

            c.         60 000 J /60 s = 1000 W

 

4.         a.         0.100 kW

            b.         400 W

            c.         1000 w = 1 kW

 

5.         I = V/R = 120/15 = 8A

            P = VI = 120(8) = 960 W

 

6.         P = I2R

            300 W = I2 (10)

            I = 5.5 A

 

7.         P = I2R

P = 0.2502(20 + 20 + 20)

            = 3.75 W

 

8.         E = Pt, where P is in W and t in seconds to obtain E in J.

a.         48 600 000 J = 48 600 kJ = 48 600 kJ (1kWh/3600 kJ) = 13.5 kWh

            b.         3 744 000 J = 3744 kJ = 3744 kJ (1kWh/3600 kJ) = 1.04 kWh

            c.         48600 J = 48.6 kJ = 48. 6 kJ (1kWh/3600 kJ) = 0.0135 kWh

            d.         666000 J = 666 kJ = 666 kJ (1kWh/3600 kJ) = 0.185 kWh

 

9.         E = Pt = 30 J/s(300 min)(60s/min) = 648 000 J = 648 kJ.

 

10.       The 2nd resistor should read 240  , not 12 to be realistic. But you don’t have to use it.

            E = VIt, where 2A is the total current.

= 120 J/C(2 C/s)(3600s/h*2h) =1728000J = 1724 kJ

 

11.       The following information is found on the back of a television:

 

 

Model SFMCL

Serial # : 181920

 

120 V

60 Hz

1.5 A

 

This television is used an average of 8 hours a day for a week.

 

How much electrical energy in kJ does this television use during these 7 days?

 

            E = VI t = 120(1.5)(7days)(8h/day)(3600s/h) = 36 288 000 J = 36 288 kJ

 

12.       A toaster is connected to a source that has a potential difference (voltage) of 110 V.  The appliance uses 990 kJ when operating for 20 minutes.

 

What current intensity (I) is flowing through this appliance when it is operating?

 

            E = VI t

            990 000 = 110(I)(20min * 60 s/min)

            I = 7.5 A

 

13.       The resistance of a heating element is 10 W and the potential difference (voltage) across its terminals is 240 V.  This element is used for 3 hours.

 

How much electric energy in kJ was used during this period?

 

I = V/R = 240/10 = 24 A

E = VIt

= 240(24)(3 h*3600s/h)

= 62 208 000 J

 

14.       The internal resistance of an electric circuit is 20 W.  The potential difference (voltage) across the terminals of this circuit is 10 V.  This circuit was used for 30 min.

 

How much energy in J did this circuit use in 30 minutes?

I = V/R = 10/20 = 0.5 A

E = VIt

= 10(0.5)(30*60)

= 9000 J

 

15.       A coffee maker is connected to a 120 V outlet.  The resistance, R, of the heating element of the coffee maker is 20 W.  This coffee maker stays on for 2 hours.

 

How much energy is used by the heating element of the coffee maker during this period?

                                   

I = V/R = 120/20 = 6.0 A

E = VIt

= 120(6)(2*3600)

= 5 184 000 J = 5184 kJ