question 1) you have one light which offers 12 Ω of resistance and has 2 a of current running...

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Question • 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with the same amount of current running through it. Find the voltage of each of the two lights.

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Page 1: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Question

• 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with the same amount of current running through it. Find the voltage of each of the two lights.

Page 2: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

• First light = 48 W·hr• Second light = 24 W·hr• 3)What if the first light is used in the

bathroom of your house and the second is used in your kitchen? Which would use more electricity?

Page 3: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Power, Electric Consumption and Efficiency

Rate of Energy UseElectrical Efficiency

Page 4: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Energy

• Defined as the ability to do work• Measured in joules (J)• Electrical energy is represented by “E”– Can be calculated with E = V x I x Δt

E= Electrical energy (J)V = voltage drop (V)I = electric current (A)Δt = time interval (s)

Page 5: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Calculating the Amount of Energy

• If the time interval is in seconds, we end up getting a large number.

• Since one joule (J) measures a very small amount of energy—enough to light a light bulb in a hundredth of a second—we use watt hour or kilowatt hour (W·hr or kW·hr)

E = V x I x Δt E= Electrical energy (W·hr)V = voltage drop (V)I = electric current (A)Δt = time interval (hr)

Page 6: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Watt is Power?

• Measured in watts (W) or kilowatts (1 kW = 1000 W)

• It measures the rate at which electrical energy is being used.

• Represented by the letter “P”• Can be solved by:

P = V x I V = Voltage dropI = Electric current

Page 7: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Energy Consumption

• Once we calculate our power, we can easily further calculate the energy consumption (EC).– Time interval in seconds = energy consumption in

joules (J). – Time interval in hours = energy consumption in

watts·hour (W·hr).

EC = P x Δt P = Power usedΔ t = time interval

Page 8: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Calculating Energy Efficiency

• Recall from the lesson on resistors:– Energy cannot be created or destroyed– It can be changed from one form to another.• No change is 100% efficient. Some energy is often

“lost” to unusable forms.

Page 9: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Calculating Energy Efficiency

• Let’s take the example of burning coal in power plants– Chemical energy in fossil

fuels would be considered our input energy (Ein)

– Electrical energy produced by the conversion is refered to as output energy (Eout)

Page 10: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Calculating Energy Efficiency

Efficiency = Energy Output

Energy Input

% Efficiency = Energy Output x 100%

Energy Input

Page 11: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Efficiency from Non-Renewable Energy Sources

Fuel Used for Generation Percentage of Energy from Fuel Available as Energy

Nuclear 20.0Petroleum 20.2Natural Gas 20.7Coal 22.2Advanced Natural Gas Turbine

26.4

Why do we use low efficiency methods when wind generators are 30 %

Page 12: Question 1) You have one light which offers 12 Ω of resistance and has 2 A of current running through it. Another light offers 6 Ω of resistance, with

Exercise With Efficiency

• An incandescent bulb produces 5 J of light energy from 100 J of electrical energy. A fluorescent bulb produces 400 J of light energy from 2000 J of electrical energy. Find the percent efficiency of each bulb and propose which is the best one to use.