Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Series

Figure 4.3.5 shows a core with a short air gap of length $l_{g}$ cut into it. There is one flux tube: the same flux $Φ$ crosses the iron (mean length $l_{c}$ , permeability $μ$ , area $S$ ) and the gap (length $l_{g}$ , permeability $μ_{0}$ , the same area $S$ when fringing is neglected). Ampère's law around the mean path has two terms on the left,

⟹ H_{c} l_{c} + H_{g} l_{g} = N I \frac{Φ l _{c}}{μ S} + \frac{Φ l _{g}}{μ _{0} S} = N I,

so the magnetic potential drops $Z Φ$ add, and with them the reluctances.

Set the gap to zero and then to 1 mm: the two drops in the bar always add up to $N I$ , and the flux follows $Φ = N I / (Z_{c} + Z_{g})$ .

InteractiveLoading simulation…

Reluctances in series

Branches carrying the same flux are in series. Their reluctances add:

Z_{eq} = Z_{c} + Z_{g} = \frac{l _{c}}{μ S} + \frac{l _{g}}{μ _{0} S}, Φ = \frac{F _{m}}{Z _{c} + Z _{g}} . (4.3.17)

(b) Its equivalent circuit. Reluctances in series. (a) The flux \Phi driven by NI crosses the iron and then the air gap: one flux tube, one flux. (b) In the equivalent circuit the iron reluctance Z_c (orange) and the gap reluctance Z_g (cyan) are in series with the mmf source F_m=NI. — **Figure 4.3.5.** (b) Its equivalent circuit. Reluctances in series. (a) The flux $Φ$ driven by $N I$ crosses the iron and then the air gap: one flux tube, one flux. (b) In the equivalent circuit the iron reluctance $Z_{c}$ (orange) and the gap reluctance $Z_{g}$ (cyan) are in series with the mmf source $F_{m} = N I$ .

Parallel

In the three-limb core of Figure 4.3.6 the flux $Φ_{1}$ driven around the left window arrives at the corner A and divides: $Φ_{2}$ goes down the central limb and $Φ_{3}$ goes round the right window. The node rule at A gives $Φ_{1} = Φ_{2} + Φ_{3}$ . Both branches start at A and end at B, so Ampère's law around the right window (which links no current) gives $Z_{2} Φ_{2} - Z_{3} Φ_{3} = 0$ : the two branches have the same magnetic potential difference $U_{A B} = Z_{2} Φ_{2} = Z_{3} Φ_{3}$ . Eliminating $Φ_{2}$ and $Φ_{3}$ ,

⟹ Φ_{1} = Φ_{2} + Φ_{3} = U_{A B} (\frac{1}{Z _{2}} + \frac{1}{Z _{3}}) U_{A B} = \frac{Z _{2} Z _{3}}{Z _{2} + Z _{3}} Φ_{1} .

Reluctances in parallel

Branches spanning the same pair of nodes are in parallel. Their reluctances combine like parallel resistors, and the whole circuit of Figure 4.3.6 has

Z_{23} = \frac{Z _{2} Z _{3}}{Z _{2} + Z _{3}}, Φ_{1} = \frac{F _{m}}{Z _{1} + Z _{23}}, Φ_{2} = \frac{Z _{3}}{Z _{2} + Z _{3}} Φ_{1}, Φ_{3} = \frac{Z _{2}}{Z _{2} + Z _{3}} Φ_{1} . (4.3.18)

The flux divides in inverse proportion to the reluctances: most of it takes the easier path.

(b) Its equivalent circuit. Reluctances in parallel. (a) At the corner A the flux \Phi_1 divides into \Phi_2 (central limb) and \Phi_3 (right window), which meet again at B. (b) Z_2 and Z_3 span the same nodes A and B, so they are in parallel, and the pair is in series with Z_1 and the source. — **Figure 4.3.6.** (b) Its equivalent circuit. Reluctances in parallel. (a) At the corner A the flux $Φ_{1}$ divides into $Φ_{2}$ (central limb) and $Φ_{3}$ (right window), which meet again at B. (b) $Z_{2}$ and $Z_{3}$ span the same nodes A and B, so they are in parallel, and the pair is in series with $Z_{1}$ and the source.

The two rules behind every magnetic circuit

The loop rule $\sum Z Φ = \sum F_{m}$ around every closed path of the circuit is $\oint_{C} H \cdot d l = N I$ written branch by branch. The node rule $\sum Φ = 0$ at every node is $\oint_{S} B \cdot d S = 0$ for a closed surface wrapped around the node. Series and parallel combinations, mesh equations and the whole of circuit theory follow from these two, as they do from Kirchhoff's two laws in electric circuits.

Recipe · Solving a magnetic circuit

Draw the mean flux path through every branch, mark the nodes where the iron branches, and give each branch its length $l$ , area $S$ and material.
Draw the equivalent circuit: a reluctance per branch, an mmf source $N I$ in the branch that carries a winding (with the direction given by the right-hand rule), and the nodes.
If the material is linear, compute $Z = l / (μ_{0} μ_{r} S)$ for every branch, with $μ_{r} = 1$ in air gaps, and solve the network with series and parallel rules or mesh equations.
If the material is nonlinear, work with $B$ and $H$ instead of $Z$ : start in the branch where $B$ or $Φ$ is given, read $H$ off the $B$ – $H$ curve, write $H l$ for the potential drop of each iron branch and $B l_{g} / μ_{0}$ for each gap, and use the loop and node rules to move to the next branch (Section 4.3.6 and Problems 4.3.1–4.3.3).
Check: every loop must balance to its $\sum N I$ , and the fluxes must balance at every node.

Quick check

Two identical outer loops of reluctance $Z_{o}$ each hang in parallel on a central limb of reluctance $Z_{c}$ that carries the winding. What reluctance does the winding see, and how does the flux divide?

$Z_{eq} = Z_{c} + Z_{o} /2$ , and by symmetry each outer loop takes half of the flux of the central limb (Home exercise 4.3.2).

4.3.4 Reluctances in series and in parallel

Series

Parallel