--- title: "No load current of single phase transformer" subject: "Electric Machine Design" chapter: 3 key: "semester_5/electric_machine_design:transformer-no-load-current" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/transformer-no-load-current site: BCT Engineering --- # No load current of single phase transformer Electric Machine Design · Chapter 3: Transformer Design - Asked in: 2069 Asar, 8 marks - Video: [VTU EMD 18EE55 M3 L9 No load current of single phase Transformer](https://www.youtube.com/watch?v=ojR5l5Lo9hQ) — Canara Engineering College Mangalore - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/transformer-no-load-current - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3 ## Question Determine the no load current of a 6600/400 V, 50 Hz, 1-ph, core-type transformer from the following data: Mean length of flux path = 270 cm; Net area of cross section of iron = 130 cm²; Maximum flux density = 1.2 Wb/m²; Specific core loss at 50 Hz and 1.2 Wb/m² = 2.1 W/kg; mmf per meter for iron at a flux density of 1.2 Wb/m² = 650 A/m; Density of iron = 7.5×10³ [kg/m³]; Take the effect of joints is equal to an air gap of length 1 mm in series with iron. ## Short answer No-load current I0 = 1.009 A (core-loss component 0.0838 A, magnetising component 1.005 A, power factor 0.083). ## Answer The no-load current $I_0$ has two components: the core-loss (active) component $I_c$, in phase with $V$, and the magnetising component $I_m$, lagging $V$ by 90°. $I_0 = \sqrt{I_c^2 + I_m^2}$. #### Core-loss component $$ \begin{aligned} \text{Volume of iron} &= l_i A_i = 2.7\times130\times10^{-4} = 0.0351\ \text{m}^3\\ \text{Weight} &= 0.0351\times7500 = 263.25\ \text{kg}\\ P_i &= 2.1\times263.25 = 552.8\ \text{W}\\ I_c &= \frac{P_i}{V_1} = \frac{552.8}{6600} = 0.0838\ \text{A} \end{aligned} $$ #### Primary turns $$ T_1 = \frac{V_1}{4.44fB_mA_i} = \frac{6600}{4.44\times50\times1.2\times0.013} = 1905.8 \approx 1906 $$ #### Magnetising component Maximum mmf needed: $$ \begin{aligned} AT_{iron} &= at\times l_i = 650\times2.7 = 1755\\ AT_{joints} &= \frac{B_m}{\mu_0}l_g = \frac{1.2}{4\pi\times10^{-7}}\times1\times10^{-3} = 954.9\\ AT_{max} &= 1755 + 954.9 = 2709.9 \end{aligned} $$ Assuming sinusoidal magnetising current, the rms value is $$ I_m = \frac{AT_{max}}{\sqrt2\,T_1} = \frac{2709.9}{\sqrt2\times1906} = 1.0054\ \text{A} $$ #### No-load current $$ \begin{aligned} I_0 &= \sqrt{I_c^2 + I_m^2} = \sqrt{0.0838^2 + 1.0054^2} = 1.0088\ \text{A}\\ \cos\phi_0 &= \frac{I_c}{I_0} = 0.0830 \end{aligned} $$ ```text Ic = 0.084 A O--------> V (reference) |\ | \ I0 = 1.009 A | \ v v Im = 1.005 A (lags V by 90 deg) ``` **Answer: no-load current ≈ 1.009 A (core-loss component 0.0838 A, magnetising component 1.005 A, no-load pf ≈ 0.083).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the two components of the no-load current? A. Core-loss component in phase with V and magnetising component lagging V by 90 degrees B. Copper-loss component and leakage component both in phase with V C. Magnetising component in phase with V and core-loss component lagging by 90 degrees D. Load component and magnetising component, 180 degrees apart Answer: A. I0 = sqrt(Ic^2 + Im^2), with Ic in phase with V and Im lagging by 90 degrees. (Key point: No-load current components) 2. How is Ic found? A. Ic = Pi/V1 where Pi = specific loss x weight of iron = 2.1 x 263.25 = 552.8 W, giving 0.0838 A B. Ic = AT/T1 = 1.0 A C. Ic = Pi/V2 with V2 = 400 V D. Ic = specific loss/V1 = 2.1/6600 Answer: A. Iron weight is volume x density = 0.0351 x 7500 = 263.25 kg, core loss 552.8 W, so Ic = 552.8/6600 = 0.0838 A. (Key point: Core-loss component) 3. How is the maximum mmf for the magnetising current found? A. Only at x li = 1755 AT; joints are ignored B. AT of iron (650 x 2.7 = 1755) plus AT of joints (Bm/mu0 x lg = 954.9) = 2709.9 C. Bm/mu0 x li = 1.9x10^6 AT D. 650 x 1 mm only Answer: B. The joints act as a 1 mm air gap needing Bm*lg/mu0 = 954.9 AT, added to the iron's at*li = 1755 AT. (Key point: Maximum mmf with joints) 4. Given T1 = 1906, what is Im and why is I0 almost equal to Im? A. Im = ATmax/T1 = 1.42 A, and Ic is larger than Im B. Im = ATmax/(sqrt2 T1) = 1.005 A, and I0 = 1.009 A since Ic = 0.084 A is tiny C. Im = ATmax x T1, and I0 = Ic D. Im = 0.084 A and I0 = 1.005 A Answer: B. The rms magnetising current is the peak AT divided by sqrt2 and the turns; the core-loss component is small, so I0 = sqrt(0.0838^2 + 1.0054^2) = 1.009 A. 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