--- title: "Per unit resistance drop of core type transformer" subject: "Electric Machine Design" chapter: 3 key: "semester_5/electric_machine_design:per-unit-resistance-drop" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/per-unit-resistance-drop site: BCT Engineering --- # Per unit resistance drop of core type transformer Electric Machine Design · Chapter 3: Transformer Design - Asked in: 2073 Shrawan, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/per-unit-resistance-drop - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3 ## Question Derive the expressions for per unit resistance drop of a core type transformer. ## Short answer ε_r = (ρ (δ_p L_mtp + δ_s L_mts)) / E_t; if equal current density δ and mean turn L_mt, ε_r = 2 ρ δ L_mt / E_t. This equals full‑load copper loss per phase divided by the phase VA rating. ## Answer Per-unit resistance drop is the full-load resistive voltage drop expressed as a fraction of rated voltage, $\varepsilon_r = I_pR_p/V_p$. It can be found from winding dimensions as below. #### Notation - $T_p$, $T_s$ = primary and secondary turns; $I_p$, $I_s$ = currents - $L_{mtp}$, $L_{mts}$ = mean lengths of turn - $a_p$, $a_s$ = conductor areas; $\delta_p$, $\delta_s$ = current densities - $\rho$ = resistivity of copper; $E_t$ = volts per turn #### Derivation Resistance of each winding: $$ r_p = \frac{\rho T_pL_{mtp}}{a_p},\qquad r_s = \frac{\rho T_sL_{mts}}{a_s} $$ Total resistance referred to primary: $$ R_p = r_p + r_s\left(\frac{T_p}{T_s}\right)^2 $$ Per-unit resistance: $$ \varepsilon_r = \frac{I_pR_p}{V_p} = \frac{I_pr_p}{V_p} + \frac{I_pr_s(T_p/T_s)^2}{V_p} $$ Since $I_pT_p = I_sT_s$ and $V_p/T_p = V_s/T_s = E_t$, the second term equals $I_sr_s/V_s$. So $$ \varepsilon_r = \frac{I_pr_p}{V_p} + \frac{I_sr_s}{V_s} $$ Now $I_pr_p = I_p\dfrac{\rho T_pL_{mtp}}{a_p} = \rho\,\delta_p\,T_pL_{mtp}$ (as $I_p/a_p = \delta_p$), and $V_p = E_tT_p$: $$ \frac{I_pr_p}{V_p} = \frac{\rho\,\delta_pL_{mtp}}{E_t},\qquad \frac{I_sr_s}{V_s} = \frac{\rho\,\delta_sL_{mts}}{E_t} $$ Therefore $$ \varepsilon_r = \frac{\rho\left(\delta_pL_{mtp} + \delta_sL_{mts}\right)}{E_t} $$ If both windings have the same current density $\delta$ and a common mean turn $L_{mt}$: $$ \varepsilon_r = \frac{2\rho\,\delta\,L_{mt}}{E_t} $$ #### Other forms Multiplying the numerator and denominator by $I_p$: $$ \varepsilon_r = \frac{I_p^2R_p}{V_pI_p} = \frac{\text{full-load copper loss (per phase)}}{\text{VA rating (per phase)}} $$ #### Example With $\rho$ = 0.021 Ω-mm²/m, $\delta$ = 2.5 A/mm², $L_{mt}$ = 0.9 m and $E_t$ = 7.5 V: $$ \varepsilon_r = \frac{2\times0.021\times2.5\times0.9}{7.5} = 0.0126\ \text{p.u.} = 1.26\% $$ #### Conclusions - $\varepsilon_r$ falls if $E_t$ is high (fewer turns) or $\delta$ is low. - A short mean turn (stepped core, compact windings) reduces both copper loss and $\varepsilon_r$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Per unit resistance drop is: A. Vp / Ip Rp B. Ip^2 Rp C. Rp / Xp D. Ip Rp / Vp Answer: D. Full-load resistive drop as a fraction of rated voltage. (Key point: Definition) 2. Total resistance referred to primary is: A. rp + rs (Tp/Ts)^2 B. rp + rs (Ts/Tp)^2 C. rp + rs D. rp rs Answer: A. Secondary resistance is referred by the turns ratio squared. (Key point: Referred resistance) 3. With equal delta and Lmt, er is: A. rho delta Lmt Et B. 2 rho delta Lmt / Et C. rho Lmt / (delta Et) D. 2 rho Et / delta Answer: B. Ip rp / Vp = rho delta Lmp / Et for each winding. (Key point: Final expression) 4. With rho 0.021, delta 2.5, Lmt 0.9 m, Et 7.5 V, er is: A. 0.126 p.u. B. 0.0063 p.u. C. 0.0126 p.u. D. 0.063 p.u. Answer: C. 2 x 0.021 x 2.5 x 0.9 / 7.5 = 0.0126. 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