EEI

Tutorial 6: Worked solutions

Electrical Engineering I · Module 92315 · Tutorial 6

Track the conserved current through each cylindrical sector, use local Ohm’s law in each material, then integrate the radial field across both layers.

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Task 10: Stationary electric flow field

The conductor is an annular sector with axial height h and angular span α. An ideal inner electrode at radius a0 feeds current through material κ1 from a0 to a1, then through material κ2 from a1 to a2 to the outer electrode. Take positive I and U12 in the outward direction from electrode 1 to electrode 2, as drawn; use α in radians.

Connected topic: Current density (§2.1.2) and local Ohm’s law (§2.2.2) determine the radial fields; resistance (§2.2.4) follows by integrating through the layered path.

Seen from above. The conductor is an annular sector of angle α and height h: κ1 from a0 to a1, κ2 from a1 to a2, between the ideal electrodes 1 (inner) and 2 (outer). The current runs radially outward; at radius ρ it crosses the area αρh.

10a: Set up the current-density field

Try first: use charge conservation at steady state and the normal-current boundary condition at the κ1/κ2 interface.

Attempted it? Open the worked steps

10a · Continuity across the layers

  1. In a stationary flow, charge does not accumulate in the conductor, so the net current through each coaxial sector surface is the same. At the interface ρ = a1, the normal is eρ: its normal current component is continuous, and the tangential electric-field components (zero here) are continuous too.

  2. The current flows radially, normal to the cylindrical interface. Therefore Jρ,1 = Jρ,2. Local Ohm’s law J = κE means the electric field generally changes at the interface:

    Eρ,2Eρ,1 = κ1κ2

  3. The ideal electrodes have κ → ∞, so their internal electric field is zero while they remain equipotential. Neglect edge spreading at the ends of the sector; the current in the two finite-conductivity layers is radial.

10b: Find J as a function of I

Try first: calculate the area normal to radial flow at radius ρ, then divide the total current by that area.

Attempted it? Open the worked steps

10b · Current through a cylindrical sector

  1. At radius ρ, the curved surface spans arc length α ρ and axial height h, so its area normal to the flow is:

    A(ρ) = αρh

  2. By symmetry the radial current density is uniform over this surface, and I = ∫ J · dA = JρA(ρ):

    J(ρ) = Iαhρ eρ,    a0 < ρ < a2

    The same expression holds on both sides of a1. It falls as 1/ρ because the current-carrying area grows with radius.

10c: Find the electric field strength

Try first: apply J = κE separately in each radial layer.

Attempted it? Open the worked steps

10c · Piecewise Ohm’s law

  1. In the first layer, a0 < ρ < a1, divide the current density by κ1:

    E1(ρ) = Iαhκ1 ρ eρ

  2. In the second layer, a1 < ρ < a2, use κ2 instead:

    E2(ρ) = Iαhκ2 ρ eρ

    At the interface the normal current remains continuous, while E jumps by the conductivity ratio found in 10a. The direction is outward for positive I.

10d: Find U12(I)

Try first: integrate the radial field along the current path, changing conductivity at the material boundary.

Attempted it? Open the worked steps

10d · Voltage drop across both layers

  1. With U12 defined as electrode-1 potential minus electrode-2 potential, it equals the outward field integral from a0 to a2. Split the integral where κ changes:

    U12 = ∫a0a1 E1,ρ dρ + ∫a1a2 E2,ρ dρ

  2. Substitute the fields and integrate dρ/ρ in each interval:

    U12(I) = Iαhκ1ln(a1a0) + Iαhκ2ln(a2a1)

    The voltage is positive for outward positive current. Each logarithm is dimensionless.

10e: Find the resistance

Try first: divide the voltage from 10d by the current, or add the two layer resistances in series.

Attempted it? Open the worked steps

10e · Series resistance of radial layers

  1. Resistance is R = U12/I. Since the current passes through κ1 and κ2 sequentially, their radial resistances add:

    R1 = 1αhκ1ln(a1a0),    R2 = 1αhκ2ln(a2a1)

  2. Therefore:

    R = 1αh[1κ1ln(a1a0) + 1κ2ln(a2a1)]

    The units are ohms: 1/(κh) has units Ω, and the angle and logarithms are dimensionless. If κ1 = κ2 = κ, the logs combine to ln(a2/a0), a useful check.

Personal worked-solution guide based on the supplied EE1 Tutorial Task sheet, written in my own words; no tutorial text or figures are reproduced. Not an official FAU publication; check course details against current material. Made by Thanate Amornjiraporn (Focus).