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Electrical Work

Electrical Work. EGR 1301: Introduction to Engineering. Electrical Work Method 1. Work (or Energy) = Force x Distance Force = magnetic force on electrons Distance = traveled in wire or resistor. +. +. I. R. V drop = I R. V batt. -. -. Voltage – What Is It?.

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Electrical Work

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  1. Electrical Work EGR 1301: Introduction to Engineering

  2. Electrical WorkMethod 1 • Work (or Energy) = Force x Distance • Force = magnetic force on electrons • Distance = traveled in wire or resistor + + I R Vdrop= IR Vbatt - -

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

  4. 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. Electrical WorkMethod 2 • Recall: Power = Energy/Time • So, or • Recall: Power = Voltage*Current (P =VI) • So, V2 E E = Pt E = VIt = P = R t t = I2Rt + + I R Vdrop= IR Vbatt - -

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

  8. 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 V C V V - - - - -

  9. 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

  10. 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. 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

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

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