1. 19-3: Electric Power Objectives:
Relate electric power to the rate at which electrical energy is converted to other forms of energy.
Calculate electric power.
Calculate the cost of running electrical appliances

2. Power and Energy in Electric Circuits
The power delivered by an electric circuit increases with both the current and the voltage. Increase either, and the power increases.
When a ball falls in a gravitational field, there is a change in gravitational potential energy. Similarly, when an amount of charge, ΔQ, moves across a potential difference, V, there is a change in electrical potential energy, ΔPE, given by ΔPE = (ΔQ)V

3. Power and Energy in Electric Circuits
Recalling that power is the rate at which energy changes, P = ΔE/Δt, we can express the electric power as follows: P = ΔE/Δt = (ΔQ)V/Δt
Knowing that the electric current is given by I = (ΔQ)/Δt allows us to write an expression for the electric power in terms of the current and voltage.

4. Power and Energy in Electric Circuits
Thus, the electric power used by a device is equal to the current times the voltage. For example, a current of 1 amp flowing across a potential difference of 1 V produces a power of 1 W.

6. Power and Energy in Electric Circuits
The equation P = IV applies to any electrical system. In the special case of a resistor, the electric power is dissipated in the form of heat and light, as shown in the figure, where the electric power dissipated in an electric space heater.

7. Power and Energy in Electric Circuits
Applying Ohm's law, V = IR, which deals with resistors, we can express the power dissipated in a resistor as follows: P = IV = I(IR) = I2R
Similarly, solving Ohm's law for the current, I = V/R, and substituting that result gives an alternative expression for the power dissipated in a resistor: P = IV = (V/R)V = V2/R
All three equations for power are valid. The first, P = IV, applies to all electrical systems. The other two (P = I2R and P = V2/R) are specific to resistors, which is why the resistance, R, appears in those equations.

8. An electric space heater is connected across a 120 V outlet
An electric space heater is connected across a 120 V outlet. The heater dissipates 1320 W of power in the form of electromagnetic radiation and heat. Calculate the resistance of the heater.
Given: ∆V = 120 V P = 1320 W
Unknown: R = ?
R = 10.9 Ω

9. Power and Energy in Electric Circuits
The power dissipated by a resistor is the result of collisions between electrons moving through the circuit and the atoms making up the resistor.
The potential difference produced by the battery causes conduction electrons to accelerate until they bounce off an atom, causing the atoms to jiggle more rapidly.
The increased kinetic energy of the atoms is reflected as an increased temperature of the resistor. After each collision, the potential difference accelerates the electrons again, and the process repeats. The result is the continuous transfer of energy from the conducting electrons to the atoms.

10. Power and Energy in Electric Circuits
The filament of an incandescent light bulb is basically a resistor inside a sealed, evacuated tube. The filament gets so hot that it glows, just like the heating coil on a stove or the coils in a space heater.

11. Power and Energy in Electric Circuits
The local electric company bills consumers for the electricity they use each month. To do this, they use a convenient unit for measuring electric energy called the kilowatt-hour.
Recall that a kilowatt is 1000 W, or equivalently, 1000 J/s. Similarly, an hour is 3600 s. Combining these results, we see that a kilowatt-hour is equal to 3.6 million joules of energy:
1 kWh = (1000 J/s)(3600 s) = 3.6 x 106 J

12. Power and Energy in Electric Circuits
The figure below shows the type of meter used to measure the electrical energy consumption of a household, as well as the typical bill.

13. How much does it cost to operate a 100
How much does it cost to operate a W light bulb for 24 h if electrical energy costs $0.080 per kW• h?
Given: Cost of energy = $0.080/kW•h P = W = kW ∆t = 24 h
Unknown: Cost to operate the light bulb for 24 h
Energy = P∆t = ( kW)(24 h) = 2.4 kW•h
Cost = (2.4 kW•h)($0.080/kW•h)
Cost = $0.19

14. Electrical energy is transferred at high potential differences to minimize energy loss
Power companies want to minimize the I2R loss and maximize the energy delivered to a consumer. This can be done by decreasing either current or resistance. Although wires have little resistance, recall that resistance is proportional to length. Even though power lines are designed to minimize resistance, some energy will be lost due to the length of the power lines.
As expressed by the equation P = I2R, energy loss is proportional to the square of the current in the wire.
For this reason, decreasing current is even more important than decreasing resistance.
Because P = I∆V, the same amount of power can be transported either at high currents and low potential differences or at low currents and high potential differences. Thus, transferring electrical energy at low currents, thereby minimizing the I2R loss, requires that electrical energy be transported at very high potential differences.
Power plants transport electrical energy at potential differences of up to 765,000 V. In your city, this potential difference is reduced by a transformer to about 4,000 V. At your home, this potential difference is reduced again to about 120 V by another transformer.

15. Assignment 19-3 Concept Review “Electric Power” Formulas:
V=IR (Ohm’s Law from 19-2)
P=IV (Be able to solve for any of the variables) This is to solve for the power used.
P= V2/R and P=I2R, when trying to find the power lost or dissipated.