--- title: "Secondary voltage, regulation, pf for max regulation" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:tf-2500-500-regulation-pf" source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-2500-500-regulation-pf site: BCT Engineering --- # Secondary voltage, regulation, pf for max regulation Electrical Machines · Chapter 2: Transformer - Asked in: 2078 Poush, 6+2 marks - Video: [Examples on Voltage Regulation of Transformer (15)](https://www.youtube.com/watch?v=hRtN2g_X-L4) — Learn Electrical Yourself - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-2500-500-regulation-pf - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question A 10 kVA, single phase transformer for 2500/500 V has $R_1 = 5.5\ \Omega$, $X_1 = 12\ \Omega$, $R_2 = 0.2\ \Omega$, $X_2 = 0.45\ \Omega$. Determine the appropriate value of secondary voltage and % voltage regulation at full load, 0.8 pf lagging, when primary applied voltage is 2000 V. Also calculate the power factor for maximum regulation. ## Short answer Refer all impedances to the secondary; equivalent series values are 0.42 Ω and 0.93 Ω. At 20 A and 0.8 pf lagging, secondary terminal voltage is about 382 V, so regulation is 4.5%. Maximum regulation occurs at cosφ = 0.412 lagging, giving VRmax = 5.1%. ## Answer **Given:** 10 kVA, 2500/500 V, $R_1 = 5.5\ \Omega$, $X_1 = 12\ \Omega$, $R_2 = 0.2\ \Omega$, $X_2 = 0.45\ \Omega$, $V_1 = 2000$ V, load pf $0.8$ lagging at full load. **Refer everything to the secondary** ($K = 500/2500 = 0.2$, $K^2 = 0.04$): $$ \begin{aligned} R_{02} &= R_2 + K^2R_1 = 0.2 + 0.04\times5.5 = 0.42\ \Omega \\ X_{02} &= X_2 + K^2X_1 = 0.45 + 0.04\times12 = 0.93\ \Omega \\ Z_{02} &= \sqrt{0.42^2 + 0.93^2} = 1.0204\ \Omega \end{aligned} $$ **Full-load current and no-load secondary voltage** $$ I_2 = \frac{10\,000}{500} = 20\ \text{A},\qquad E_2 = KV_1 = 0.2\times2000 = 400\ \text{V} $$ Since the exciting current is neglected, the secondary no-load voltage is 400 V with the 2000 V supply. **Secondary terminal voltage** (with $\cos\phi = 0.8$, $\sin\phi = 0.6$). Using $V_2$ as reference: $$ \begin{aligned} E_2^2 &= (V_2 + I_2R_{02}\cos\phi + I_2X_{02}\sin\phi)^2 + (I_2X_{02}\cos\phi - I_2R_{02}\sin\phi)^2 \\ I_2R_{02}\cos\phi + I_2X_{02}\sin\phi &= 20(0.336 + 0.558) = 17.88\ \text{V} \\ I_2X_{02}\cos\phi - I_2R_{02}\sin\phi &= 20(0.744 - 0.252) = 9.84\ \text{V} \\ 400^2 &= (V_2 + 17.88)^2 + 9.84^2 \Rightarrow V_2 = 382.0\ \text{V} \end{aligned} $$ **Percentage regulation** $$ \text{VR} = \frac{400 - 382.0}{400}\times100 = 4.50\% $$ (The approximate formula gives $\dfrac{17.88}{400}\times100 = 4.47\%$.) **Power factor for maximum regulation** Regulation is greatest when $\tan\phi = X_{02}/R_{02}$, i.e. $$ \cos\phi = \frac{R_{02}}{Z_{02}} = \frac{0.42}{1.0204} = 0.4116\ \text{lagging} $$ The maximum regulation is $\dfrac{I_2Z_{02}}{E_2}\times100 = \dfrac{20\times1.0204}{400}\times100 = 5.10\%$. **Answer: $V_2 \approx 382$ V; regulation $\approx 4.5\%$; pf for maximum regulation $= 0.412$ lagging (VR$_{max} = 5.1\%$)** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is R02 found for the secondary side? A. R2 + R1/K B. R2 + K^2 R1 with K = 0.2 C. R2 + K R1 D. R2 + R1 / K^2 Answer: B. Primary resistance referred to secondary is multiplied by K^2 = 0.04, giving 0.42 ohm. (Key point: Referring to secondary) 2. With 2000 V applied, what is the secondary no-load voltage? A. 500 V B. 2000 V C. 400 V D. 382 V Answer: C. E2 = K V1 = 0.2 x 2000 = 400 V. (Key point: No-load voltage) 3. At what power factor is regulation maximum? A. cos phi = R02/Z02 = 0.412 lagging B. Unity C. cos phi = X02/Z02 = 0.911 lagging D. 0.8 lagging Answer: A. Regulation is maximum when tan phi = X02/R02, giving cos phi = R02/Z02. 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