6 hours · 6 marks
A single-phase transformer has two magnetically coupled windings wound on a laminated soft-iron (or silicon-steel) core, with no electrical connection between them (except in an auto-transformer). Main parts:
Core type: windings surround the limbs; more copper, less iron — common for high voltage.
Shell type: iron surrounds the windings; more iron, better short-circuit strength — common for low-voltage distribution.
Based on Faraday’s law of mutual induction. An alternating voltage $V_1$ applied to the primary produces alternating flux $\Phi = \Phi_m\sin\omega t$ in the core. This flux links both windings and induces EMFs:
$$e = -N\frac{d\phi}{dt}$$Energy is transferred magnetically from primary to secondary with (ideally) no change in frequency and nearly equal power (neglecting losses).
Instantaneous flux: $\phi = \Phi_m\sin\omega t$, $\omega=2\pi f$.
$$e = -N\frac{d\phi}{dt} = -N\Phi_m\omega\cos\omega t = N\Phi_m\,2\pi f\sin(\omega t-90^\circ)$$Maximum EMF: $E_m = 2\pi f\,N\Phi_m$. RMS value:
$$E = \frac{E_m}{\sqrt{2}} = 4.44\,f\,N\,\Phi_m$$For primary and secondary:
$$E_1 = 4.44\,f\,N_1\,\Phi_m,\qquad E_2 = 4.44\,f\,N_2\,\Phi_m$$Turns ratio / transformation ratio:
$$K = \frac{E_2}{E_1} = \frac{N_2}{N_1} \approx \frac{V_2}{V_1}$$Secondary open. Primary draws a small no-load current $I_0$ (2–5% of full-load current), resolving into:
Induced $E_1 \approx V_1$ (no $I R$ drop at no load, neglecting $I_0 R_1$). Flux $\Phi_m = E_1/(4.44 f N_1)$ is fixed by supply voltage and frequency.
When load is connected, secondary current $I_2$ flows. Secondary mmf $I_2 N_2$ tends to demagnetize the core. Primary current increases by a load component $I_2'$ so that:
$$I_1' N_1 = I_2 N_2 \quad\Rightarrow\quad I_1' = K I_2$$Total primary current: $\vec{I}_1 = \vec{I}_0 + \vec{I}_1'$.
From Kirchhoff’s law on the primary (steady state):
$$\vec{V}_1 = \vec{I}_1 R_1 + j\vec{I}_1 X_1 + (-\vec{E}_1)$$At no load, $I_0$ is very small, so $I_0 R_1$ and $I_0 X_1$ drops are negligible → $E_1 \approx V_1$.
On load, primary current rises, but the increase in $IR$ and $IX$ drops is typically only a few percent of $V_1$. Hence $E_1$ (and therefore $\Phi_m = E_1/(4.44 f N_1)$) remains essentially constant — main flux is nearly independent of load.
Physically: any attempt by secondary mmf to reduce flux is immediately opposed by an equal and opposite primary load mmf ($I_1' N_1 = I_2 N_2$), so net ampere-turns (and flux) stay almost the same as at no load.
A practical transformer is represented by resistances, leakage reactances, and a magnetizing branch:
Referred to secondary (divide primary quantities by $K^2$ for impedance, by $K$ for voltage/current as appropriate):
$$R_1' = K^2 R_1,\quad X_1' = K^2 X_1$$ $$R_{02} = R_2 + R_1',\quad X_{02} = X_2 + X_1',\quad Z_{02} = \sqrt{R_{02}^2 + X_{02}^2}$$Referred to primary:
$$R_2' = \frac{R_2}{K^2},\quad X_2' = \frac{X_2}{K^2},\quad R_{01}=R_1+R_2',\quad X_{01}=X_1+X_2'$$Approximate equivalent circuit: shunt branch moved to the supply terminals (valid because $I_0 \ll I_1$).
Usually performed on the LV side; HV left open. Measure $V_0$, $I_0$, $W_0$.
$$W_0 = \text{iron (core) loss } P_i \text{ (full rated voltage)}$$ $$\cos\phi_0 = \frac{W_0}{V_0 I_0},\quad I_w = I_0\cos\phi_0,\quad I_m = I_0\sin\phi_0$$ $$R_0 = \frac{V_0}{I_w},\qquad X_m = \frac{V_0}{I_m}$$Gives shunt-branch parameters and constant iron loss.
Usually on HV side; LV shorted. Apply reduced voltage $V_{sc}$ so that full-load current flows. Measure $V_{sc}$, $I_{sc}$, $W_{sc}$.
$$W_{sc} = \text{full-load copper loss } P_{\mathrm{cu,fl}}$$ $$Z_{01} = \frac{V_{sc}}{I_{sc}},\quad R_{01} = \frac{W_{sc}}{I_{sc}^2},\quad X_{01} = \sqrt{Z_{01}^2 - R_{01}^2}$$Iron loss during SC is negligible (flux very small at reduced voltage). Gives series parameters of the equivalent circuit.
Change in secondary terminal voltage from no-load to full-load, at constant primary voltage, expressed as % of full-load (or rated) secondary voltage:
$$\mathrm{VR} = \frac{E_2 - V_2}{V_2}\times 100\% \approx \frac{I_2 R_{02}\cos\phi \pm I_2 X_{02}\sin\phi}{V_2}\times 100\%$$$+$ for lagging pf, $−$ for leading pf. At unity pf, $\mathrm{VR}\approx I_2 R_{02}/V_2$.
| Loss | Nature | How obtained |
|---|---|---|
| Iron / core ($P_i$) | Hysteresis + eddy; nearly constant | OC test ($W_0$) |
| Copper ($P_{\mathrm{cu}}$) | $I^2 R$; varies as (load)$^2$ | SC test ($W_{sc}$ at FL) |
$P_{\mathrm{cu}} = x^2 P_{\mathrm{cu,fl}}$ where $x$ = load fraction (p.u. of rated kVA).
Differentiate $\eta$ w.r.t. $x$ and set to zero:
$$P_i = x^2 P_{\mathrm{cu,fl}} \quad\Rightarrow\quad x = \sqrt{\frac{P_i}{P_{\mathrm{cu,fl}}}}$$i.e. iron loss = copper loss at that load. Load kVA for $\eta_{\max} = x\times$ rated kVA.
Single winding acts as both primary and secondary; part of the winding is common. Conductively connected.
Used for: starting induction motors, interconnecting grids of close voltages, laboratory variable AC (Variac).
$K=1$ (or near 1) two-winding transformer providing galvanic isolation between supply and load — safety, noise reduction, breaking ground loops. No intentional voltage change; focus is isolation.
Q. A 150 kVA transformer has iron loss of 700 W and full-load copper loss of 1800 W. Find the losses and efficiency at 0.8 pf when the transformer is 25% overloaded.
Given: $S=150\,\mathrm{kVA}$, $P_i=700\,\mathrm{W}$, $P_{\mathrm{cu,fl}}=1800\,\mathrm{W}$, overload 25% → $x=1.25$, $\cos\phi=0.8$.
Copper loss at 1.25× load:
$$P_{\mathrm{cu}} = x^2 P_{\mathrm{cu,fl}} = (1.25)^2\times 1800 = 1.5625\times 1800 = 2812.5\,\mathrm{W}$$Iron loss (constant): $P_i = 700\,\mathrm{W}$
Total losses: $2812.5 + 700 = 3512.5\,\mathrm{W}$
Output power:
$$P_{\mathrm{out}} = x\,S\cos\phi = 1.25\times 150\times 0.8 = 150\,\mathrm{kW} = 150000\,\mathrm{W}$$Efficiency:
$$\eta = \frac{150000}{150000+3512.5}\times 100\% = \frac{150000}{153512.5}\times 100\% = \mathbf{97.71\%}$$Answer: Total losses = 3512.5 W; efficiency = 97.71%.
Q. A 20 kVA, 2200/220 V, 50 Hz single-phase transformer gave the following test results:
Draw the equivalent circuit referred to the secondary.
Step 1 — Shunt branch (from OC on LV = secondary side already):
$$\cos\phi_0 = \frac{150}{220\times 4} = 0.1705,\quad I_w = 4\times 0.1705 = 0.682\,\mathrm{A},\quad I_m = \sqrt{4^2-0.682^2}=3.941\,\mathrm{A}$$ $$R_{02}' = R_0 = \frac{220}{0.682} = 322.6\,\Omega,\qquad X_{m2} = \frac{220}{3.941} = 55.8\,\Omega$$(These $R_0$, $X_m$ are already referred to secondary / LV.)
Step 2 — Series branch from SC on HV (primary):
$$Z_{01} = \frac{100}{9} = 11.11\,\Omega,\quad R_{01} = \frac{400}{9^2} = 4.938\,\Omega,\quad X_{01} = \sqrt{11.11^2-4.938^2}=9.95\,\Omega$$Refer to secondary: $K = V_2/V_1 = 220/2200 = 0.1$
$$R_{02} = K^2 R_{01} = (0.1)^2\times 4.938 = 0.0494\,\Omega$$ $$X_{02} = K^2 X_{01} = 0.01\times 9.95 = 0.0995\,\Omega$$Equivalent circuit referred to secondary:
Iron loss from OC = 150 W (constant). Full-load copper loss = 400 W (from SC at ~rated current: rated $I_1=20000/2200=9.09\,\mathrm{A}\approx 9\,\mathrm{A}$).
For the 150 kVA transformer of Example 1 ($P_i=700\,\mathrm{W}$, $P_{\mathrm{cu,fl}}=1800\,\mathrm{W}$), find kVA at maximum efficiency and $\eta_{\max}$ at unity pf.
$$x = \sqrt{\frac{P_i}{P_{\mathrm{cu,fl}}}} = \sqrt{\frac{700}{1800}} = 0.6236$$ $$\text{kVA for }\eta_{\max} = 0.6236\times 150 = 93.54\,\mathrm{kVA}$$ $$P_{\mathrm{cu}}=P_i=700\,\mathrm{W},\quad P_{\mathrm{out}}=93.54\times 1\times 1000=93540\,\mathrm{W}$$ $$\eta_{\max}=\frac{93540}{93540+700+700}\times 100\% = \mathbf{98.53\%}$$