egr 1301 electrical work egr 1301: introduction to engineering
TRANSCRIPT
EGR 1301
Electrical Work
EGR 1301: Introduction to Engineering
EGR 1301
Electrical WorkMethod 1
• Work (or Energy) = Force x Distance Force = magnetic force on electrons Distance = traveled in wire or resistor
Vbatt RI
Vdrop = IR
+
-
+
-
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)
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.
EGR 1301
Electrical WorkMethod 2
• Recall: Power = Energy/Time So, or
• Recall: Power = Voltage*Current (P =VI) So,
Vbatt RI
Vdrop = IR
+
-
+
-
tE
P = E = Pt
RV2
E = VIt = t = I2Rt
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
EGR 1301
Capacitors
• Energy is stored in electric field between the plates.
• Recall:• From method 1: E = QV
V C
+
-
VQ
C =
++ ++
- - - -
Conductor
Insulator
Stored Charge
or Q = VC
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
+
-
++ ++
- - - -VVV
VQE QdVdE
VVEVCdVQdVdE
000
2
2
1CVE
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 = ½CV2 = ½(50 μF)(10 V)2
= ½(50x10-6 F)(10 V)2 = 2.5x10-3 J = 2.5 mJ
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?
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 = ½CV2 = ½(100x10-6 F)(250 V)2 = 3.125 J 1 bulb: E = Pt Ebulbs = 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
EGR 1301
Inductor
• Energy is stored in magnetic field inside the coil.
• Similar to the capacitor, except using current instead of voltage
2
2
1LIE
L
+ -V
I