1. EGR 1301 Electrical Work EGR 1301: Introduction to Engineering

2. EGR 1301 Electrical Work Method 1 Work (or Energy) = Force x Distance  Force = magnetic force on electrons  Distance = traveled in wire or resistor V batt R I V drop = I R + - + -

3. EGR 1301 Voltage – What Is It? Energy per unit charge of current 1 Volt = Or V = E/Q So E = VQ 1 Coulomb(charge) 1 Joule (energy)

4. EGR 1301 Example 1: Battery A charge of Q = 50 C flows through a 12 V battery.  a) How much energy is imparted to the charge?  E = VQ = (12 V)(50 C) = 240 J  b) Where does the energy come from?  The electro-chemical reactions in the battery.

5. EGR 1301 Electrical Work Method 2 Recall: Power = Energy/Time  So, or Recall: Power = Voltage*Current (P =V I )  So, V batt R I V drop = I R + - + - t E P = E = Pt R V2V2 E = V I t = t = I 2 R t

6. EGR 1301 Example 2: Water Heater How much energy does a hot water heater use if it draws 10 A from a 120 V wall outlet for 1 hour?  I = 10 A, V = 120 V, t = 1 hr = 3600 s  E = VIt = (120 V)(10 A)(3600 s) = 4.32 MJ  Units check: V*A*s = J/C * C/s * s = J

7. EGR 1301 Capacitors Energy is stored in electric field between the plates. Recall: From method 1: E = QV VC + - V Q C = ++++ ---- Conductor Insulator Stored Charge or Q = VC

8. EGR 1301 Capacitors From previous slide: E = QV and Q = VC Charge builds up on either side of capacitor Each bit of charge requires more energy V C + - ++++ ---- V V V

9. EGR 1301 Example 3: Capacitor A 50 μF capacitor is charged to 10 V. What energy is stored?  C = 50 μF, V = 10 V  E = ½CV 2 = ½(50 μF)(10 V) 2  = ½(50x10 -6 F)(10 V) 2  = 2.5x10 -3 J = 2.5 mJ

10. EGR 1301 Example 3: Camera Flash Assume a light bulb and a camera flash give the same light per unit energy. The camera flash has a 100 μF capacitor, charged to 250 V. How many 60 W light bulbs is this energy usage equivalent to if the camera discharges in t = 0.01 s?

11. EGR 1301 Example 3: Camera Flash How many 60W light bulbs is this energy usage equivalent to if the camera discharges in t = 0.01 s?  C = 100 μF, V = 250 V, t = 0.01 s  E = ½CV 2 = ½(100x10 -6 F)(250 V) 2 = 3.125 J  1 bulb: E = Pt  E bulbs = NPt = N(60 W)(0.01 s) = 0.6*N J  0.6*N J = 3.125 J  N = 3.125/0.6 = 5.21 bulbs

12. EGR 1301 Inductor Energy is stored in magnetic field inside the coil. Similar to the capacitor, except using current instead of voltage L +-V I