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Capacitor Examples

C. C. C. C. Capacitor Examples. 2 C. C /2. d/4. 3d/4. Magnitude. rate at which net positive charges move across a cross sectional surface. Units: [I] = C/s = A (ampere).

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Capacitor Examples

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  1. C C C C Capacitor Examples 2C C/2 d/4 3d/4

  2. Magnitude rate at which net positive charges move across a cross sectional surface Units:[I] = C/s = A (ampere) Current is ascalar, signed quantity, whose sign corresponds to the direction ofmotion of netpositive charges by convention Electric Current Current = charges in motion J = current density (vector) in A/m²

  3. A Microscopic View of Electric Current in Conductor All charges move with some velocity ve random motion with high speeds (O(106)m/s) but with adrift in a certain direction on average if E is present • thermal energy • scattering off each other, defects, ions, … Drift velocity vdis orders of magnitudes less than the actual velocity of charges,

  4. where n=carrier density Current and Drift Velocity in Conductor Drift velocityvdis orders of magnitudes less than the actual velocity of charges. In the following condition: I = 1.0A, copper: n ~1029atoms/m3 1mm radius wire. Vd~0.01mm/s

  5. Current-Potential (I-V) characteristic of a device may or may not obey Ohm’s Law: or V  IR with R constant (ohms) Resistance gas in fluorescent tube diode tungsten wire Ohm’s Law

  6. resistivity A L R (in ) resistance Resistance and Resitivity for Ohmic Material

  7. R I V constant ROhm’s Law Resistance Resistance (definition)

  8. R I V Warm up There are 2x1014 electrons across a resistor in 10 seconds. What is the current through the resistor? a) 3.2mA b) 1.6 mA c) 3.2 A d) 1.6A e) 3.2 mA Note: e = 1.6x10-19 C

  9. Usually T0 is 293K (room temp.) • Usually  > 0 (ρ increases as T ) Temperature Dependence of Resistivity Copper

  10. Transfers energy to the atoms of the solid (to vibrate), i.e., Joule heating. Electric Current and Joule Heating • Free electrons in a conductor gains kinetic energy due to an externally applied E. • Scattering from the atomic ions of the metal and other electrons quickly leads to a steady state with a constant current I.

  11. Energy lost by Q is So, Power dissipation = rate of decrease of U = Energy in Electric Circuits • Steady current means a constant amount of charge Q flows past any given cross section during time t, where I= Q / t. I I a b a b ∆Q ∆Q => heat

  12. EMF – Electromotive Force • An EMF device is a charge pump that can maintain a potential difference across two terminals by doing work on the charges when necessary. Examples:battery, fuel cell, electric generator, solar cell, fuel cell, thermopile, … • Converts energy (chemical, mechanical, solar, thermal, …) into electrical energy. • Within the EMF device, positive charges are lifted from lower to higher potential. • If work dW is required to lift charge dq, EMF

  13. load terminal voltage internal resistance Internal Resistance of a Battery

  14. Energy conservation Work done by battery is equal to energy dissipated in resistor dW > i2Rdt or  > iR=V Energy Conservation A circuit consists of an ideal battery (B) with emf , a resistor R, and two connecting wires of negligible resistance. • Idealbattery:no internal energy dissipation • Realbattery: internal energy dissipation exists

  15. R I V Lecture quiz A There are1014 electrons across a resistor with potential drop of 3.2mV in 10 seconds. What is the resistance of the resistor? a) 2.0 Ω b) 1.0 Ω c) 2.5 Ω d) 3.0 Ω e) 4.0 Ω Note: e = 1.6x10-19 C

  16. R I V Lecture quiz B There are 1014 electrons across a resistor of resistance 1.0Ω in 10 seconds. What is the potential drop across the resistor? a)3.2 mV b)8.0 V c)2.5 V d)1.6 mV e)1.6mV Note: e = 1.6x10-19 C

  17. R I V Lecture quiz C The potential drop is 6.4mV across a resistor of resistance 1.0Ω. How many electrons enter the wire in 10 seconds? a)3.2×1019 b)8.0×1015 c)2.5×1012 d)4.0x1014 e)1.6×1019 Note: e = 1.6x10-19 C

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