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Prof. David R. Jackson ECE Dept.

ECE 2317 Applied Electricity and Magnetism. Spring 2014. Prof. David R. Jackson ECE Dept. Notes 26. KCL Law. i 1. i N. Wires meet at a “node”. i 2. i 4. i 3. Note: The “node” can be a single point or a larger region. KCL Law (cont.). +. v. -. A single node is considered.

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Prof. David R. Jackson ECE Dept.

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  1. ECE 2317 Applied Electricity and Magnetism Spring 2014 Prof. David R. Jackson ECE Dept. Notes 26

  2. KCL Law i1 iN Wires meet at a “node” i2 i4 i3 Note: The “node” can be a single point or a larger region.

  3. KCL Law (cont.) + v - A single node is considered. A= surface area of the “node.” A +Q C is the stray capacitance between the “node” and ground. C Ground The node forms the top plate of a stray capacitor.

  4. KCL Law (cont.) Two cases for which the KCL law is valid: • In “steady state” (no time change) 2) As area of node A0

  5. KCL Law (cont.) In general, the KCL will be accurate if the size of the “node” is small compared with the wavelength 0. Currents enter a node at some frequency f. Node

  6. KCL Law (cont.) S Example where the KCL is not valid Open-circuited transmission line (phasor domain) Zg I(z) + Z0 V(z) - z

  7. KCL Law (cont.) S Open-circuited transmission line Current enters this region (“node”) but does not leave. Zg I(z) Z0 z

  8. KCL Law (cont.) General volume (3D) form of KCL equation: (valid for D.C. currents) The total current flowing out must be zero: whatever flows in must flow out. S

  9. KCL Law (Differential Form) J V To obtain the differential form of the KCL law for static (D.C.) currents, start with the definition of divergence: For the right-hand side: Hence (valid for D.C. currents)

  10. Important Current Formulas These two formulas hold in general (not only at DC). Ohm’s Law (This is an experimental law that was introduced earlier in the semester.) Charge-Current Formula (This was derived earlier in the semester.)

  11. Resistor Formula L x A J + V - Note: The electric field is constant since the current must be uniform (KCL law). A long narrow resistor: I Solve for V from the last equation: Hence we have We also have that

  12. Joule’s Law v = charge density from electrons in conduction band. v A B v - - - - - - - - V E Conducting body • W= work (energy) given to a small volume of charge as it moves inside the conductor from point A to point B. This goes to heat! (There is no acceleration of charges in steady state, as this would violate the KCL law.)

  13. Joule’s Law (cont.) The total power dissipated is then We can also write

  14. Power Dissipation by Resistor Passive sign convention labeling Resistor V + - x A I L Hence Note: The passive sign convention applies to the VI formula.

  15. RC Analogy EC ER B B A A Insulating material Conducting material I - - VAB + VAB + “Rproblem” “Cproblem” Goal: Assuming we know how to solve the C problem (i.e., find C), can we solve the R problem (find R)?

  16. RC Analogy (cont.) EC ER B B A A I VAB + - VAB - + “Rproblem” “Cproblem” Theorem: EC = ER (same field in both problems)

  17. RC Analogy (cont.) Proof “Rproblem” “Cproblem” • Same D. E. since (r) = (r) • Same B. C. since the same voltage is applied EC = ER Hence, (They must be the same from uniqueness of the solution.)

  18. RC Analogy (cont.) EC B A + - VAB Use Hence “C Problem”

  19. RC Analogy (cont.) ER B A I + - VAB Use Hence “R Problem”

  20. RC Analogy (cont.) Compare: Recall that Hence

  21. RC Analogy Recipe for calculating resistance: • Calculate the capacitance of the corresponding C problem. • Replace  everywhere with  to obtain G. • Take the reciprocal to obtain R. In equation form:

  22. RC Formula This is a special case: A homogeneousmedium of conductivity  surrounds the two conductors. Hence, or

  23. Example Find R A h Method #1 (RC analogy or “recipe”) C problem: A h

  24. Example Find R A h C problem: Method #2 (RC formula) A h

  25. Example Find the resistance A  2 h2 h1  1 Note: We cannot use the RC formula, since there is more than one region (not a single conductivity).

  26. Example C problem: A 2 h2 h1 1

  27. Example (cont.)

  28. Example (cont.) A  2 h2 h1  1 Hence, we have

  29. Lossy Capacitor A [m2] Q + i (t) + + + + + + + + + + + + + + + + + + v(t) E h J - - - - - - - - - - - - - - - - - - - - - - - - - - - i (t) + v (t) R - This is modeled by a parallel equivalent circuit (proof on next slides). Note: The time constant is C

  30. Lossy Capacitor (cont.) A [m2] Q + i (t) + + + + + + + + + + + + + + + + + + v(t) E h J x - - - - - - - - - - - - - - - - - - - - - - - - - - - Total current entering top (A) plate: Therefore,

  31. Lossy Capacitor (cont.) A [m2] Q + i (t) + + + + + + + + + + + + + + + + + + v(t) E h J x - - - - - - - - - - - - - - - - - - - - - - - - - - -

  32. Lossy Capacitor (cont.) i (t) + v (t) R - A [m2] Q + i (t) + + + + + + + + + + + + + + + + + + v(t) E h J x - - - - - - - - - - - - - - - - - - - - - - - - - - - C This is the KCL equation for a resistor in parallel with a capacitor.

  33. Lossy Capacitor (cont.) A [m2] Q + i (t) + + + + + + + + + + + + + + + + + + v(t) E h J x - - - - - - - - - - - - - - - - - - - - - - - - - - - Ampere’s law: We can also write Conduction current (density) Displacement current (density) Displacement current Conduction current

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