--- title: "Exciting current waveform with non-linear B-H core" subject: "Electrical Machines I" chapter: 1 key: "semester_4/electrical_machines_i:nonlinear-bh-exciting-current" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/nonlinear-bh-exciting-current site: BCT Engineering --- # Exciting current waveform with non-linear B-H core Electrical Machines I · Chapter 1: Magnetic Circuits and Induction - Asked in: 2080 Chaitra, 4+4 marks - Video: [Magnetic Circuits - Sinusoidal Excitation](https://www.youtube.com/watch?v=bV5L4dEZiq8) — Energy Conversion Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/nonlinear-bh-exciting-current - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1 ## Question For a non-linear characteristic of BH relationship of a transformer core, what would be the nature of current for sinusoidal excitation? Explain with necessary expression and figures. ## Short answer For sinusoidal voltage, flux is forced sinusoidal. The nonlinear B–H curve makes the exciting current non‑sinusoidal and peaked, containing odd harmonics (mainly 3rd). The current can be expressed as a Fourier series, with its fundamental split into magnetizing and core‑loss components. ## Answer For a sinusoidal supply voltage the core flux is forced to be **sinusoidal**; because the B–H (φ–i) curve is non-linear (saturation and hysteresis), the **exciting current is non-sinusoidal**. It is peaky and contains odd harmonics, mainly the 3rd. #### Why the flux is sinusoidal Neglecting winding resistance and leakage, $$ v = V_m \cos\omega t = N\frac{d\phi}{dt} \Rightarrow \phi = \frac{V_m}{N\omega}\sin\omega t = \phi_m \sin\omega t $$ So the flux follows the voltage waveform, whatever the core is like. The current needed for each value of flux is then read from the φ–i curve. #### Shape of the exciting current (graphical method) ```text phi-i curve waveforms vs time phi phi (sine) i0 (peaky) | ___ | _ ^ | / | / \ / \ | / |/ \ __./ \.__ ---+------ i0 ---+-------\----------- t /| \_/ _/ | ``` Steps: for each instant take φ from the sine wave, find the matching i on the φ–i curve, and plot i against time. 1. **Saturation only (no hysteresis):** near the peak of φ the core is saturated, so a small extra flux needs a large current. The current is **peaked**, symmetric about the flux peak, and **in phase with φ**. It contains the fundamental plus large 3rd, 5th, … harmonics. 2. **With hysteresis:** the rising and falling branches differ, so the current wave is also **unsymmetrical** and its zero crossings come **before** those of the flux. The current leads the flux slightly. #### Expression and components The exciting current can be written as a Fourier series of odd harmonics: $$ i_0(t) = I_1\sin(\omega t + \theta_1) + I_3\sin(3\omega t + \theta_3) + I_5\sin(5\omega t + \theta_5) + \dots $$ The fundamental component is split into: - **Magnetising current** $I_m = I_0 \sin\phi_0$, in phase with φ (lags V by 90°); it sets up the flux. - **Core-loss current** $I_c = I_0\cos\phi_0$, in phase with V; it supplies hysteresis and eddy losses, $P_c = V I_c$. $$ I_0 = \sqrt{I_m^2 + I_c^2} $$ #### Consequences - The 3rd harmonic is typically 30–40% of the fundamental in a transformer worked near the knee. - In three-phase banks, a delta winding or a neutral path is provided for these 3rd-harmonic currents; otherwise the flux and voltages become distorted. - Because $I_0$ is only 2–5% of rated current, the distortion is usually neglected and an equivalent sine wave is used in the equivalent circuit. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a sinusoidal supply voltage (neglecting resistance and leakage), the core flux is: A. Peaky B. Sinusoidal C. Square D. Triangular Answer: B. v = N dphi/dt forces phi to follow the voltage as a sine wave whatever the core is like. (Key point: Flux waveform) 2. Because the phi-i curve is non-linear, the exciting current is: A. Sinusoidal B. Non-sinusoidal and peaky, with odd harmonics C. Constant D. Sinusoidal with even harmonics Answer: B. Saturation makes the current peaky; it contains odd harmonics, mainly the third. (Key point: Current waveform) 3. How is the exciting current wave obtained graphically? A. For each instant take phi from the sine wave and read i from the phi-i curve B. Take i from the voltage wave directly C. Multiply phi by the turns D. Differentiate the flux wave Answer: A. Each flux value has a matching current on the phi-i curve, which is plotted against time. (Key point: Graphical method) 4. Including hysteresis, the current wave becomes: A. Symmetrical and in phase with flux B. Unsymmetrical, with the zero crossings before those of the flux C. Smaller with no harmonics D. Lagging the flux by 90 degrees Answer: B. The rising and falling branches differ, so the current leads the flux slightly (core-loss component). 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