Prof. David R. Jackson Dept. of ECE

1. ECE 3318 Applied Electricity and Magnetism Spring 2019 Prof. David R. Jackson Dept. of ECE Notes 27 DC Currents

2. KCL Law Wires meet at a “node” Note: The “node” can be a single point or a larger region. A proof of the KCL law is given next.

3. KCL Law (cont.) A single node is considered. A= surface area of the “node.” C is the stray capacitance between the “node” and ground. + - 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 KCLlaw 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.) Example where the KCL is not valid Open-circuited transmission line (ECE 3317) + - + - (phasor domain)

7. KCL Law (cont.) Open-circuited transmission line Current enters this shaded region (“node”) but does not leave. + - + -

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

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

12. Joule’s Law (Please see one of the appendices for a derivation.) Conducting body The power dissipated inside the body as heat is:

13. Power Dissipation by Resistor Resistor Passive sign convention labeling + - Assume a uniform current. Hence Note: The passive sign convention applies to the VI formula.

14. RC Analogy Insulating material Conducting material EC ER B B A A I - VAB + - (same conductors) + 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)?

15. RC Analogy (Please see one of the appendices for a derivation.) 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 symbolic form:

16. RC Formula This is a special case: A homogeneousmedium of conductivity  surrounds the two conductors (there is only one value of ).  = conductivity in resistor problem  = permittivity in capacitor problem The resistance R of the resistor problem is related to the capacitance C of the capacitor problem as follows:

17. Example Find R Method #1 (RC analogy or “recipe”) C problem: Hence

18. Example Find R C problem: Method #2 (RC formula)

19. Example Find the resistance Note: We cannot use the RC formula, since there is more than one region (not a single conductivity).

20. Example (cont.) C problem:

21. Example (cont.)

22. Example (cont.) Hence, we have

23. Lossy Capacitor + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - - This is modeled by a parallel equivalent circuit (the proof is in one of the appendices). Note: The time constant is + -

24. Appendix Derivation of Joule’s Law

25. Joule’s Law v = charge density from electrons in conduction band - - - - - - - - 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 cause current to change along the conductor, violating the KCL law.)

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

27. Appendix Derivation of RC Analogy

28. RC Analogy Insulating material Conducting material EC ER B B A A I - VAB + - + VAB (same conductors) “Rproblem” “Cproblem”

29. RC Analogy (cont.) I + - - + (same conductors) “Rproblem” “Cproblem” Theorem: EC = ER (same field in both problems)

30. RC Analogy (cont.) Proof of “theorem” “Rproblem” “Cproblem” • Same differential equation since (r) = (r) • Same B. C. since the same voltage is applied Hence, (They must be the same function from the uniqueness of the solution to the differential equation.)

31. RC Analogy (cont.) Use - + Hence “C Problem”

32. RC Analogy (cont.) I Use + - Hence “R Problem”

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

34. RC Formula This is a special case: A homogeneousmedium of conductivity  surrounds the two conductors (there is only one value of ). Hence, or

35. Appendix Equivalent Circuit for Lossy Capacitor

36. LossyCapacitor Derivation of equivalent circuit + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - - Therefore, Total (net) current entering top (A) plate:

37. Lossy Capacitor (cont.) + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - -

38. Lossy Capacitor (cont.) + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - - + - This is the KCL equation for a resistor in parallel with a capacitor.

39. Lossy Capacitor (cont.) + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - - Ampere’s law: Note on displacement current: We can also write the current as Conduction current (density) Displacement current (density) Displacement current Conduction current