EMF Equation and Turns Ratio
Transformer EMF Equation
E = 4.44 \, f \, N \, \Phi_m
| Symbol | Description | Unit |
|---|---|---|
| E | RMS EMF induced | V |
| f | Supply frequency | Hz |
| N | Number of turns | turns |
| Φm | Maximum flux in core | Wb |
Worked example
A single-phase transformer has 500 primary turns, connected to 230V 50Hz supply. Find the maximum core flux.
Given: E1=230V, f=50Hz, N1=500
- E = 4.44 * f * N * Φm
- Φm = E / (4.44 * f * N)
- = 230 / (4.44 * 50 * 500)
- = 230 / 111,000
- = 2.07 × 10^{-3} Wb
- = 2.07 mWb
Answer: Φm = 2.07 mWb
Turns Ratio
a = \frac{N_1}{N_2} = \frac{E_1}{E_2} = \frac{V_1}{V_2} = \frac{I_2}{I_1}
| Symbol | Description | Unit |
|---|---|---|
| a | Turns ratio (primary to secondary) | — |
| N1, N2 | Primary and secondary turns | turns |
| I1, I2 | Primary and secondary currents | A |
Worked example
A 230V/115V transformer has N1=800 turns. Find N2 and the current ratio when I1=5A.
Given: V1=230V, V2=115V, N1=800, I1=5A
- a = V1/V2 = 230/115 = 2
- N2 = N1/a = 800/2 = 400 turns
- I2/I1 = N1/N2 = a → I2 = a * I1 = 2 * 5 = 10A
Answer: N2 = 400, I2 = 10A
Impedance Transformation
Z_1 = a^2 Z_2, \quad Z_2' = a^2 Z_2
| Symbol | Description | Unit |
|---|---|---|
| Z1 | Impedance reflected to primary side | Ω |
| Z2 | Actual secondary-side impedance | Ω |
| a² | Square of turns ratio | — |
Worked example
A transformer has a=5. A 10Ω load on the secondary is reflected to the primary. Find Z_reflected.
Given: a=5, Z2=10Ω
- Z_reflected = a^2 * Z2
- = 5^2 * 10
- = 25 * 10
- = 250Ω
Answer: Z_reflected to primary = 250Ω
Equivalent Circuit Parameters
Approximate Equivalent Circuit Referred to Primary
R_{eq1} = R_1 + a^2 R_2, \quad X_{eq1} = X_1 + a^2 X_2
| Symbol | Description | Unit |
|---|---|---|
| R1, R2 | Primary and secondary winding resistances | Ω |
| X1, X2 | Primary and secondary leakage reactances | Ω |
| Req1, Xeq1 | Total resistance and reactance referred to primary | Ω |
Worked example
A transformer: R1=0.5Ω, X1=1Ω, R2=0.02Ω, X2=0.04Ω, a=5. Find Req1 and Xeq1.
Given: R1=0.5, X1=1, R2=0.02, X2=0.04, a=5
- Req1 = R1 + a^2*R2 = 0.5 + 25*0.02 = 0.5 + 0.5 = 1.0Ω
- Xeq1 = X1 + a^2*X2 = 1 + 25*0.04 = 1 + 1.0 = 2.0Ω
- Zeq1 = sqrt(Req1^2 + Xeq1^2) = sqrt(1+4) = sqrt(5) = 2.24Ω
Answer: Req1 = 1.0Ω, Xeq1 = 2.0Ω, |Zeq1| = 2.24Ω
Open Circuit Test (Finds Core Loss Parameters)
R_c = \frac{V_0^2}{P_0}, \quad X_m = \frac{V_0}{I_m}, \quad I_m = \sqrt{I_0^2 - I_c^2}
| Symbol | Description | Unit |
|---|---|---|
| V0 | Rated voltage applied (LV side) | V |
| P0 | No-load power loss (core loss) | W |
| I0 | No-load current | A |
| Rc | Core loss resistance | Ω |
| Xm | Magnetising reactance | Ω |
Worked example
OC test on LV side: V0=230V, I0=1.2A, P0=100W. Find Rc and Xm.
Given: V0=230V, I0=1.2A, P0=100W
- Rc = V0^2 / P0 = 230^2 / 100 = 52,900 / 100 = 529Ω
- Power factor of no-load: cosφ0 = P0/(V0*I0) = 100/(230*1.2) = 100/276 = 0.362
- Ic = I0 * cosφ0 = 1.2 * 0.362 = 0.435A
- Im = sqrt(I0^2 - Ic^2) = sqrt(1.44 - 0.189) = sqrt(1.251) = 1.118A
- Xm = V0/Im = 230/1.118 = 205.7Ω
Answer: Rc = 529Ω, Xm = 205.7Ω
Short Circuit Test (Finds Series Impedance)
Z_{eq} = \frac{V_{sc}}{I_{sc}}, \quad R_{eq} = \frac{P_{sc}}{I_{sc}^2}, \quad X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2}
| Symbol | Description | Unit |
|---|---|---|
| V_sc | Voltage at short circuit test (rated current) | V |
| I_sc | Rated current during SC test | A |
| P_sc | Power input during SC test (copper loss) | W |
Worked example
SC test on HV side (rated current): V_sc=22V, I_sc=10A, P_sc=150W. Find Zeq, Req, Xeq referred to HV side.
Given: V_sc=22V, I_sc=10A, P_sc=150W
- Zeq = V_sc/I_sc = 22/10 = 2.2Ω
- Req = P_sc/I_sc^2 = 150/100 = 1.5Ω
- Xeq = sqrt(Zeq^2 - Req^2) = sqrt(2.2^2 - 1.5^2) = sqrt(4.84 - 2.25) = sqrt(2.59) = 1.61Ω
Answer: Zeq=2.2Ω, Req=1.5Ω, Xeq=1.61Ω (referred to HV)
Voltage Regulation
Voltage Regulation (Exact)
\%VR = \frac{V_{2,no-load} - V_{2,full-load}}{V_{2,full-load}} \times 100
| Symbol | Description | Unit |
|---|---|---|
| V2,nl | Secondary no-load voltage | V |
| V2,fl | Secondary full-load voltage | V |
Worked example
A transformer secondary gives 230V at no load and 220V at full load. Find % VR.
Given: V2_nl = 230V, V2_fl = 220V
- %VR = (230 - 220)/220 * 100
- = 10/220 * 100
- = 4.55%
Answer: %VR = 4.55%
Approximate Voltage Regulation
\%VR \approx \epsilon_R \cos\phi \pm \epsilon_X \sin\phi
| Symbol | Description | Unit |
|---|---|---|
| εR | Per-unit resistance drop = IReq/V | pu |
| εX | Per-unit reactance drop = IXeq/V | pu |
| cosφ | Load power factor | — |
| + for lag | + sign for lagging load, - for leading | — |
Worked example
A transformer has εR=0.02 pu, εX=0.04 pu. Load pf=0.8 lagging. Find approximate %VR.
Given: εR=0.02, εX=0.04, cosφ=0.8, sinφ=0.6
- For lagging load: %VR = εR*cosφ + εX*sinφ
- = 0.02*0.8 + 0.04*0.6
- = 0.016 + 0.024
- = 0.04 pu = 4%
Answer: %VR = 4% (lagging load)
Efficiency
Transformer Efficiency
\eta = \frac{x \cdot S \cdot \cos\phi}{x \cdot S \cdot \cos\phi + P_i + x^2 P_c} \times 100\%
| Symbol | Description | Unit |
|---|---|---|
| x | Fraction of full load (0 to 1) | pu |
| S | Full-load VA rating | VA |
| Pi | Iron (core) loss, constant | W |
| Pc | Full-load copper loss | W |
| cosφ | Power factor of load | — |
Worked example
A 50kVA transformer: Pi=200W, Pc (full load)=500W, pf=0.9. Find η at full load.
Given: S=50000VA, Pi=200W, Pc=500W, cosφ=0.9, x=1
- Output power = x*S*cosφ = 1*50000*0.9 = 45000W
- Total losses = Pi + x^2*Pc = 200 + 1*500 = 700W
- η = 45000/(45000+700) * 100 = 45000/45700 * 100 = 98.47%
Answer: η = 98.47% at full load, pf=0.9
Condition for Maximum Efficiency
P_i = x^2 P_c \Rightarrow x_{m} = \sqrt{\frac{P_i}{P_c}}
| Symbol | Description | Unit |
|---|---|---|
| xm | Fraction of full load at maximum efficiency | pu |
Worked example
A transformer: Pi=200W, Pc=500W. At what load fraction does maximum efficiency occur?
Given: Pi=200W, Pc=500W
- xm = sqrt(Pi/Pc) = sqrt(200/500)
- = sqrt(0.4)
- = 0.632
- So maximum efficiency occurs at 63.2% of full load
Answer: xm = 0.632 (63.2% of full load)
All-Day Efficiency
\eta_{all-day} = \frac{\text{Energy output (kWh)}}{\text{Energy output + Total energy losses}} \times 100\%
| Symbol | Description | Unit |
|---|---|---|
| Energy output | kWh output over 24 hours | kWh |
| Total losses | Iron losses × 24h + copper losses × hours at each load | kWh |
Worked example
A distribution transformer: Pi=100W. Operates 8h at full load (10kVA, pf=0.8, Pc_fl=250W) and 16h at no load. Find all-day efficiency.
Given: Pi=100W, S=10kVA, cosφ=0.8, Pc_fl=250W, 8h full load, 16h no load
- Energy output = 10000*0.8*8 = 64,000Wh = 64kWh
- Iron loss energy = 0.1kW * 24h = 2.4kWh
- Copper loss energy = 0.25kW * 8h = 2kWh (only when loaded)
- Total loss energy = 2.4 + 2.0 = 4.4kWh
- η_all-day = 64/(64+4.4) * 100 = 64/68.4 * 100 = 93.57%
Answer: η_all-day = 93.57%
Per-Unit System
Per-Unit Quantities
Z_{pu} = \frac{Z_{actual}}{Z_{base}}, \quad Z_{base} = \frac{V_{base}^2}{S_{base}}
| Symbol | Description | Unit |
|---|---|---|
| Z_pu | Per-unit impedance | pu |
| V_base | Base voltage | V |
| S_base | Base apparent power | VA |
Worked example
A 100kVA, 11kV/400V transformer has Zeq=2.5Ω referred to LV side. Find Z_pu on LV base.
Given: S_base=100,000VA, V_base=400V, Z_actual=2.5Ω
- Z_base = V_base^2 / S_base = 400^2 / 100,000 = 160,000/100,000 = 1.6Ω
- Z_pu = Z_actual / Z_base = 2.5 / 1.6 = 1.5625 pu
- This is a high leakage impedance — typical for a current-limiting transformer
Answer: Z_pu = 1.5625 pu (on LV base)
Change of Base Formula
Z_{pu,new} = Z_{pu,old} \times \frac{S_{base,new}}{S_{base,old}} \times \left(\frac{V_{base,old}}{V_{base,new}}\right)^2
| Symbol | Description | Unit |
|---|---|---|
| Z_pu,new | Per-unit impedance on new base | pu |
| S_base,new | New MVA base | VA |
| V_base,new | New voltage base | V |
Worked example
Transformer Z_pu=0.05 pu on 100kVA, 11kV base. Convert to 200kVA, 11kV base.
Given: Z_old=0.05pu, S_old=100kVA, S_new=200kVA, V same
- Z_pu,new = Z_pu,old * (S_new/S_old) * (V_old/V_new)^2
- = 0.05 * (200/100) * (1)^2
- = 0.05 * 2 = 0.10 pu
Answer: Z_pu,new = 0.10 pu on 200kVA base
Three-Phase Transformer Connections
Turns Ratio for Three-Phase Transformers
a = \frac{N_1}{N_2} = \frac{V_{1L}/\sqrt{3}}{V_{2L}/\sqrt{3}} \text{ (Y-Y)}, \quad a = \frac{V_{1L}/\sqrt{3}}{V_{2L}} \text{ (Y-Δ)}
| Symbol | Description | Unit |
|---|---|---|
| V1L, V2L | Primary and secondary line voltages | V |
Worked example
A Y-Δ transformer: V1L=11kV (Y), V2L=415V (Δ). Find the turns ratio of each single-phase unit.
Given: V1L=11000V (star), V2L=415V (delta)
- Primary phase voltage (star): V1P = 11000/sqrt(3) = 6350V
- Secondary phase voltage (delta): V2P = V2L = 415V
- Turns ratio a = V1P/V2P = 6350/415 = 15.3
Answer: Turns ratio a = 15.3 per single-phase unit
Scott Connection (3-phase to 2-phase)
\text{Teaser turns} = \frac{\sqrt{3}}{2} N, \quad \text{Main turns} = N
| Symbol | Description | Unit |
|---|---|---|
| N | Number of turns in main transformer | turns |
Worked example
A Scott connection uses main transformer with 1000 turns on the primary. How many turns does the teaser transformer need?
Given: N_main = 1000 turns
- Teaser turns = (sqrt(3)/2) * N_main
- = 0.866 * 1000
- = 866 turns
- The centre tap is at 500 turns from the main transformer
Answer: Teaser primary turns = 866
Quick reference
| Formula | Expression |
|---|---|
| EMF Equation | E = 4.44·f·N·Φm |
| Turns Ratio | a = N1/N2 = V1/V2 = I2/I1 |
| Impedance Transformation | Z_reflected = a²·Z2 |
| Req referred to primary | Req1 = R1 + a²R2 |
| OC Test: Rc | Rc = V0²/P0 |
| OC Test: Xm | Xm = V0/Im; Im=√(I0²-Ic²) |
| SC Test: Zeq | Zeq = Vsc/Isc |
| SC Test: Req | Req = Psc/Isc² |
| Efficiency at load x | η = x·S·cosφ / (x·S·cosφ + Pi + x²Pc) |
| Max efficiency condition | Pi = x²·Pc → xm = √(Pi/Pc) |
| Voltage Regulation | %VR = (V_nl - V_fl)/V_fl × 100 |
| Approx %VR | εR·cosφ ± εX·sinφ |
| Per-unit base impedance | Zbase = Vbase²/Sbase |
| Change of base | Zpu_new = Zpu_old × (Snew/Sold) × (Vold/Vnew)² |
| Delta-Star turns ratio | a = (V1L/√3)/V2L for Y-Δ |
Exam tips
- GATE repeatedly tests the OC and SC test analysis — the OC test gives core loss parameters (Rc, Xm) performed on the LV side, while the SC test gives series impedance (Req, Xeq) performed on the HV side at rated current.
- The condition for maximum efficiency is iron loss equals copper loss (Pi = x²Pc) — examiners often give Pi and Pc and ask for the load at which η_max occurs; use xm = √(Pi/Pc).
- Voltage regulation sign matters — lagging loads give positive (poor) regulation and leading loads can give negative regulation; the approximate formula εR·cosφ ± εX·sinφ with + for lag is frequently tested.
- In per-unit calculations, always convert all quantities to the same base before performing any analysis; the change-of-base formula is essential when manufacturers specify impedance on their own nameplate rating.
- All-day efficiency is specifically relevant to distribution transformers and tests whether you account for iron losses over 24 hours even at no load — a common trap is forgetting to add no-load iron loss for the unloaded hours.