Air-core reactor field/circuit studies solve parallel coil currents and linked magnetic fields together. [1] Equal DC resistance alone cannot establish current sharing: different positions and lead paths change self and mutual inductances as well as resistance.
Label every parallel branch from terminal to terminal. For a linear fixed layout, write v = R i + L di/dt as a branch-vector equation, including self and mutual terms. Here v is in volts, i in amperes, R in ohms, L in henries and t in seconds. Branches connected to common terminals share terminal voltage, not necessarily current.
Compare the complete branch paths
| Path item | Evidence | Possible imbalance |
|---|---|---|
| Conductor resistance | Length, section, joints and temperature | Resistive voltage-drop mismatch |
| Coil position | Self/mutual inductance matrix | Unequal linked flux |
| Transposition | Actual sequence of positions | Incomplete averaging of branch exposure |
| Terminal leads | Length, routing and nearby conductors | Asymmetry outside the nominal winding |
Solve current sharing on the installed frequency basis
For sinusoidal operation, with ω in rad/s, Z = R + jωL. Solve the coupled branch equations with common-terminal constraints and total current. Check individual currents and losses; dropping mutual terms changes the model. [2]
A transposition can equalize exposure when it gives branches comparable time or length in the relevant positions. Verify the complete path; a winding transposition may leave asymmetric leads. Its adequacy also depends on the field geometry, not solely on equal conductor length.
Retain the complete branch path through the terminals when comparing a proposed transposition; equal conductor length does not alone equalize linked flux.
Compare branch current with unequal inductance
Assumptions: two parallel branches at 80 °C each have R = 0.020 Ω; at 50 Hz their L₁ = 1.0 mH, L₂ = 1.2 mH and signed M = +0.8 mH. Total current is 220∠0° A RMS. Equal terminal voltages give I₁ = 220[R + jω(L₂−M)]/[2R + jω(L₁+L₂−2M)]. Magnitudes are 145.28 A and 75.28 A; the phasors sum to 220 A. Against an assumed 130 A branch limit, branch 1 fails by 15.28 A. Setting M = 0 falsely predicts 119.97 A and 100.04 A. Hold the sharing claim and verify the installed coupled matrix. [2]
References
- J. Wu et al., Thermal Simulation and Analysis of Dry-Type Air-Core Reactors Based on Multi-Physics Coupling, Energies 16, 7456, 2023. DOI: 10.3390/en16217456.
- U.S. Department of Energy, DOE-HDBK-1011/3-92, Electrical Science, Volume 3, June 1992.