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Electrodynamics

Electrodynamics. Overview HRW p592>. What is electrodynamics?. Previously we have been studying the nature of stationary charges, aka electrostatics . Electrodynamics = study of charges (q) in motion. What causes charges to flow?. Analogy - Think of a garden hose –

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Electrodynamics

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  1. Electrodynamics Overview HRW p592>

  2. What is electrodynamics? • Previously we have been studying the nature of stationary charges, aka electrostatics. • Electrodynamics = study of charges (q) in motion

  3. What causes charges to flow? • Analogy - Think of a garden hose – • Water will flow if there is a pressure differencebetween the 2 ends.

  4. What causes charges to flow? • In electrical models, the garden hose is replaced by a circuit comprising components of wires, loads and sources, and water is replaced by charges. • Charges will flow if there is a potential difference(or voltage drop) between 2 ends of the circuit. • Solids – electrons flow • Fluids – ions & electrons flow

  5. Materials • 2 types • Conductors • Electrons are free to move • Ex. • Copper, aluminum • Insulators • Electrons are tightly bound to the nucleus • Ex. • Plastic, rubber, wood (dry)

  6. Current • The movement of charges (q) past any point in a circuit in a period of time (t) is called the current. • Symbol for current • I • Unit for current • Amperes (amps – A) • Ex. 3 amps is written 3A • Formula • I = charge/time • I = q/t • 1 amp = 1 coulomb/sec

  7. Practice - Current • How much current is flowing in a circuit if 5 x 10-3C of charge (q) pass a point every 2 seconds (t)? • Solve I = q/t for I • I = 5 x 10-3/2 • I = 2.5 x 10-3 amps • I = 2.5 mA (milliamps)

  8. Potential Difference (V or E) • Think of potential difference as an “electric pump” pushing electrons around a circuit. • Units of potential difference • Volts (V) • Sources of voltage • Dry/wet cell battery (chemical energy) • Generator (mechanical energy)

  9. Resistance (R) Visually • Water hose analogy – a restriction in the hose, causing energy to be used. • Think of the model as a “pipe” as shown: Small R – Big I Big R – Small I

  10. Resistance (R) • A “load” in the circuit, causing the electrons to lose energy as they work through the segment. • A bi-product of resistance is usually heat, such as • Toaster, kettle, electric radiator • Units of resistance • Ohms (Ω) • Resistance depends on • Material resistivity (ρ) • Length (l) • Cross-sectional area (A) • R = ρl/A

  11. Ohm’s Law – most important! • Relates potential difference, current and resistance as: • potential difference = current * resistance, or • Volts = amps * ohms • V = I*R, or • I = V/R, or • R = V/I • For a given voltage, • Resistance (R) is inversely proportional to current (I) • R α 1/I

  12. Practice - Current • How much current (I) flows through a circuit containing a 6 V battery and a 33 Ω resistance? • Solve V = IR for I • I = V/R • I = 6/33 • I = 0.182 A, or • I = 182 mA

  13. Practice – Ohm’s Law • What is the potential difference (V) across a “load” if 2 amps of current (I) flows through the 5Ω (R) load? • Solve for V = IR • V = 2 x 5 • V = 10 volts

  14. Current types • 2 types • Direct current (DC) • Charges move in 1 direction only • Alternating current (AC) • Charges move in 1 direction, then change direction on a regular basis (frequency) • USA • ~120VAC, 60Hz “power” • Other places • ~240VAC, 50Hz “power”

  15. Surviving Electricity! • Current kills, not voltage • Conductors – easy current movement • Insulators – difficult current movement • Thresholds for current (varies by person) • 1 mA can be felt • 5 mA can be painful • 10 mA can cause spasms • 20 mA can cause loss of muscle control • 70 mA is lethal Electric chair: 5-10A @ ~2000V

  16. Surviving Electricity! • Body resistance is fundamental to controlling current into the body – the higher the better! • Dry body > 500,000Ω • Wet body > 100Ω • Household examples (V = 110V) • Given I = V/R • Dry: I = 110/500,000 = 0.22 mA – OK! • Wet: I = 110/100 = 1.1 A - LETHAL !

  17. Capacitance • A capacitor is a device that stores charge on a conductor, separated by an insulator called a dielectric and hence can provide electrical potential energy (PE)

  18. Examples of Capacitors

  19. Capacitance • Letter symbol • C • Formula • C = charge/voltage = coulomb/volt • C = q/V • Unit • Farad (F) • 1 farad = 1 coulomb/volt • 1F is very large, so practical capacitances are of the order of micro (μ, m) (10-6) and pico (p) (10-12) farads • Ex. 4700 mfd (= 4700*10-6 F)

  20. Capacitor Characteristics • Examine the demo capacitor for • Polarity (+/-) • Capacitance value (ex. 4700 µF) • Working voltage (ex. 35V)

  21. Practice – Capacitance • A camera flash unit releases 4.5 x 10-3 C of charge that was stored in a 500 μF capacitor. What is the potential difference (voltage) across the capacitor plates inside the flash unit? • Solve C = q/V for V • V = q/C • V = 4.5 x 10-3/500 x 10-6 • V = 9V

  22. Potential Energy in Capacitors • Capacitors store charge and thus potential energy according to the following formula: • PE = ½ * CV2 where • PE = energy measured in joules (J) • C = capacitance in farads (F) • V = potential difference in volts (V)

  23. Power • Definition • “Rate of doing work” • In electrical systems • Power (P) = work/time = w/t = V(q/t) = VI • P = VI (or P = V2/R, or P = I2R, since V = IR) • Unit • Watt (= volt-amp) • Watt = joule/sec

  24. Examples of power ratings • Microwave – 800W • Electric kettle – 1300W • Electric mixer – 150W • Coffee maker – 600W • Hair dryer – 1500W • A/C unit – 2500W • Values will vary! • How much current for each device (V=120V)?

  25. Practice – Power • How much current is drawn when a 1500W hair dryer is plugged into a 120V household circuit? • Solve P = VI for I • I = P/V • I = 1500/120 • I = 12.5 amps (A)

  26. Power Company charging • Power companies charge consumers in units of: • Kilowatt-hours (kwh) • Power (kw) x time (hours) • Multiplied by a $ charge • Ex $.10 per kwh • kwh = unit of energy/work (not power!)

  27. Practice – Domestic Electricity • If Dominion Power charges $0.1 (aka 10c) per kWh, what will be the charge for 2 x 300W computers, 3 x 60W lights, and 1 x 2500W a/c unit running for 8 hours? • Solve $ = power (kW) x time(hours) x $ • $ = (2x300 + 3x60 + 2500)/1000 x 8 x 0.1 • $ = 3.28 x 8 x 0.1 • $ = $2.62

  28. Next • DC Circuit Analysis

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