--- title: "Condition for minimum cost transformer design" subject: "Electric Machine Design" chapter: 3 key: "semester_5/electric_machine_design:transformer-minimum-cost-condition" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/transformer-minimum-cost-condition site: BCT Engineering --- # Condition for minimum cost transformer design Electric Machine Design · Chapter 3: Transformer Design - Asked in: 2071 Shrawan, 8 marks - Video: [5EE4-05 Lecture 4 (Part 1) - Unit 2 - EMD- Design for minimum cost of transformer - Gauri Shankar](https://www.youtube.com/watch?v=rzkgKLNmb9w) — MITRC LECTURE - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/transformer-minimum-cost-condition - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3 ## Question Find the condition for designing a transformer in minimum cost. ## Short answer Minimum total active‑material cost occurs when cost of iron equals cost of copper (c_i G_i = c_c G_c). This follows from minimizing C_t = k₁ φ_m + k₂ AT with φ_m·AT constant, giving φ_m/AT = (2 c_c ρ_c L_mt B_m) / (c_i ρ_i l_i δ). ## Answer A transformer has minimum total cost of active material (iron + copper) when the **cost of iron equals the cost of copper**. #### Derivation Let - $\phi_m$ = maximum flux, $AT$ = ampere-turns per limb - $c_i$, $c_c$ = cost per kg of iron and copper - $\rho_i$, $\rho_c$ = densities; $l_i$ = mean length of flux path; $L_{mt}$ = mean length of turn - $B_m$, $\delta$ = flux density and current density (fixed by losses and heating) For a single-phase transformer: $$ Q = 4.44f\phi_m\,(IT)\times10^{-3} = 4.44f\,\phi_m\,AT\times10^{-3} $$ so for a given rating the product $\phi_m \cdot AT$ is constant: $$ \phi_m\cdot AT = \frac{Q\times10^3}{4.44f} = \text{constant} = C $$ Weight of iron $G_i = \rho_il_iA_i = \rho_il_i\dfrac{\phi_m}{B_m}$. Copper area (both windings) $= \dfrac{2AT}{\delta}$, so weight of copper $G_c = \rho_cL_{mt}\dfrac{2AT}{\delta}$. Total cost: $$ C_t = c_iG_i + c_cG_c = \underbrace{\frac{c_i\rho_il_i}{B_m}}_{k_1}\phi_m + \underbrace{\frac{2c_c\rho_cL_{mt}}{\delta}}_{k_2}AT $$ Put $AT = C/\phi_m$: $$ C_t = k_1\phi_m + \frac{k_2C}{\phi_m} $$ For minimum cost: $$ \begin{aligned} \frac{dC_t}{d\phi_m} &= k_1 - \frac{k_2C}{\phi_m^2} = 0\\ k_1\phi_m &= \frac{k_2C}{\phi_m} = k_2\,AT \end{aligned} $$ that is, $$ c_iG_i = c_cG_c \quad\Longrightarrow\quad \text{cost of iron} = \text{cost of copper} $$ #### Optimum ratio of magnetic to electric loading From $k_1\phi_m = k_2AT$: $$ r = \frac{\phi_m}{AT} = \frac{k_2}{k_1} = \frac{2c_c\rho_cL_{mt}B_m}{c_i\rho_il_i\delta} $$ This ratio fixes the voltage per turn, $E_t = K\sqrt{Q}$ with $K = \sqrt{4.44fr\times10^3}$. #### Notes - In the same way, **minimum total loss** (maximum efficiency) at a load occurs when iron loss = copper loss. - In practice designers use the "minimum cost" ratio only as a guide, since loss capitalisation, regulation and cooling also matter. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A transformer has minimum cost of active material when: A. Cost of iron is double copper B. Iron weight equals copper weight C. Iron loss is zero D. Cost of iron equals cost of copper Answer: D. Setting dCt/dphi = 0 gives k1 phi = k2 AT, i.e. ci Gi = cc Gc. (Key point: Minimum cost condition) 2. For a given rating, which product is constant? A. phi_m x AT B. phi_m + AT C. phi_m / AT D. Bm x delta Answer: A. Q = 4.44 f phi_m AT x 10^-3. (Key point: Constant product) 3. After substituting AT = C/phi_m, total cost is: A. k1 phi x k2 C B. k1 phi + k2 C / phi C. k1/phi + k2 phi D. k1 phi^2 Answer: B. Differentiating this sum gives the minimum. (Key point: Cost expression) 4. Minimum total loss (maximum efficiency) occurs when: A. Iron loss is zero B. Copper loss is zero C. Iron loss equals copper loss D. Iron loss is double copper loss Answer: C. The notes state this same-style condition for loss. 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