Prof. David R. Jackson ECE Dept. Fall 2014 Notes 27 ECE 2317 Applied Electricity and Magnetism 1

KCL Law Wires meet at a “node” i2i2 i3i3 iNiN i4i4 i1i1 2 Note: The “node” can be a single point or a larger region. A proof of the KCL law is given next.

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

KCL Law (cont.) 1)In “steady state” (no time change) 2) As area of node A  0 Two cases for which the KCL law is valid: 4

KCL Law (cont.) In general, the KCL law will be accurate if the size of the “node” is small compared with the wavelength 0. f 0 60 Hz5000 [km] 1 kHz300 [km] 1 MHz300 [m] 1 GHz30 [cm] 5 Node Currents enter a node at some frequency f.

KCL Law (cont.) 6 Example where the KCL is not valid Open-circuited transmission line (phasor domain) ZgZg Z0Z0 + - V(z)V(z) I(z)I(z) z + -

KCL Law (cont.) 7 Open-circuited transmission line Current enters this region (“node”) but does not leave. ZgZg Z0Z0 I(z)I(z) z + -

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

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

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

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

Joule’s Law  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.) 12 VV A B vv E - v Conducting body  v = charge density from electrons in conduction band

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

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

RC Analogy Goal: Assuming we know how to solve the C problem (i.e., find C ), can we solve the R problem (find R )? “ C problem” A + - V AB B ECEC “ R problem” 15 A - V AB B ERER I + Insulating material Conducting material (same conductors)

RC Analogy (cont.) Theorem: E C = E R (same field in both problems) “ C problem” A + - V AB B ECEC “ R problem” 16 A + - V AB B ERER I (same conductors)

E C = E R “ C problem” “ R problem”  Same differential equation since  ( r ) =  ( r )  Same B. C. since the same voltage is applied RC Analogy (cont.) Proof Hence, 17 (They must be the same function from the uniqueness of the solution to the differential equation.)

“ C Problem” RC Analogy (cont.) Hence Use A + - V AB B ECEC 18

RC Analogy (cont.) Hence “ R Problem” Use A + - V AB B ERER I 19

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

RC Analogy Recipe for calculating resistance: 1)Calculate the capacitance of the corresponding C problem. 2)Replace  everywhere with  to obtain G. 3)Take the reciprocal to obtain R. In equation form: 21

This is a special case: A homogeneous medium of conductivity  surrounds the two conductors (there is only one value of  ). RC Formula Hence, or 22

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

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

Example Find the resistance Note: We cannot use the RC formula, since there is more than one region (not a single conductivity). 25 h1h1 h2h2  2 2  1 1 A

Example C problem: 26 h1h1 h2h2  2 2  1 1 A

Example (cont.) 27

Hence, we have Example (cont.) 28 h1h1 h2h2  2 2  1 1 A

Lossy Capacitor C v (t) R i (t) + - This is modeled by a parallel equivalent circuit (proof on next slides). A [ m 2 ] Q h E v(t)v(t) + - i (t) J Note: The time constant is 29

30 Lossy Capacitor (cont.) Therefore, A [ m 2 ] Q h E v(t)v(t) + - i (t) J x Total current entering top ( A ) plate: Derivation of equivalent circuit

31 Lossy Capacitor (cont.) A [ m 2 ] Q h E v(t)v(t) + - i (t) J x

32 Lossy Capacitor (cont.) A [ m 2 ] Q h E v(t)v(t) + - i (t) J x This is the KCL equation for a resistor in parallel with a capacitor. C v (t) R i (t) + -

33 Lossy Capacitor (cont.) We can also write A [ m 2 ] Q h E v(t)v(t) + - i (t) J x Conduction current Displacement current Ampere’s law: Conduction current (density) Displacement current (density)