1 / 12

Download Presentation
## Electric Energy

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**Electric Energy**Electric energy is made available by the flow of electric charge through a conductor. It is useful because it can easily be transformed into other types of energy. For example, motors transform electric energy into mechanical energy, and electric heaters, stoves and hair dryers transform electric energy into thermal energy.**In a light bulb**the filament gets so hot that it glows and produces visible light. 10% of the electric energy is transformed into light and the rest into thermal energy. What’s really going on when you turn on the lights is that there are many collisions occurring between the moving electrons and the atoms of the wire. The electrons transfer some of their kinetic energy to the atoms when they collide and thus, the temperature of the atoms increases. This thermal energy is then radiated as light.**Electric Power**Electric power is the rate at which energy is transformed. The energy transformed when a charge Q moves through a voltage V is QV. Thus, P = power (W or J/s) Q = charge (C) V = voltage (V) t = time (s) P = energy transformed = QV time t However, the charge per second (Q/t) is simply the electric current (I). We can therefore change the formula to read: P = power (W or J/s) I = current (C/s or A) V = voltage (V) P = IV**Electric Power**Ex: What is the power transformed by a device where 15.0 A current passes through 12.0 V? P = IV P = (15.0 A)(12.0 V) P = 180 W = 1.80 x 102 W**Electric Power**The rate of energy transformation in a resistance R can be writing using Ohm’s Law (V=IR) in two other ways: P = power (W) I = current (A) R = resistance (Ω) P = I(IR) = I2R P = (V/R)V = V2/R However, these two formulas only apply to resistors, whereas P = IV can apply to any device.**Electric Power**Ex: Determine the electric power of 150 A current in 3.5 Ω of resistance. P = I2R P = (150 A)2(3.5 Ω) P = 78750 W = 7.9 x 104 W**Electric Power**Ex: Calculate the resistance of a 40. W headlight designed for 12 V. P = V2/R R = V2/P R = (12 V)2/(40.W) R = 3.6 Ω**Ex: A 34 A current runs through a 25 V hair dryer. What is**the power transformed by the hair dryer? P = IV P = (34 A)(25 V) P = 850 W 850 W**Ex: Calculate the resistance of a device with a 15.0 A**current and a power of 635 W. R = P/I2 R = (635 W)/(15.0 A)2 R = 2.82 Ω 2.82 Ω**Ex: A 65 C current moves through a potential difference of**250 V in 4.0 seconds. Determine the power transformed. P = QV/t P = (65 C)(250 V)/(4.0 s) P = 65000 W = 6.5 x 104 W 6.5 x 10 4 W**Ex: An electric heater draws 15.0 A on a 120 V line. If it**operates 3.0 hours per day and the electric company charges 10.5 cents per kWh, how much does it cost per month (30 days)? P = IV P = (15.0 A)(120 V) P = 1800 W = 1.80 kW (3.0 h/d)(30 d) = 90 h (1.80 kW)(90 h)($ .105) = $17 $17**Ex: A bolt of lighting transfers 109 J of energy across a**potential difference of 5x107 V during a time interval of .2 s. What is the power delivered by the lightning? E = QV Q = V/E Q = (5x107 V)/(109 J) = 20 C I = Q/t I = (20 C)/(.02 s) = 100 A P = IV P = (100 A)(5x107 V) = 5000000000 W = 5 GW 5 GW

More Related