Transformers 1

LessonAbout 2 h1 of 2 in Transformers

Transformers 1

Transformer is a static electrical machine that transfers electrical energy between two or more circuits through electromagnetic induction without a change in frequency.

1. Induced EMF Equation

  • Primary Induced EMF: voltage induced in primary winding turns
  • Secondary Induced EMF: voltage induced in secondary winding turns
  • EMF Per Turn: induced voltage divided by number of turns

Induced EMF Equation

\[ E = 4.44 N f \Phi_m \]
\[ e_t = \dfrac{E}{N} = 4.44 f \Phi_m \]
\[ \Phi_m = \beta_m A \]

Where: \(E\) = rms induced voltage (V), \(N\) = number of winding turns, \(f\) = frequency (Hz), \(\Phi_m\) = peak magnetic flux (Wb), \(e_t\) = induced emf per turn (V/turn), \(\beta_m\) = peak flux density (T), \(A\) = core cross-sectional area (\(\text{m}^2\))

Example. Calculate the total magnetic flux in a \(60\ \text{Hz}\) transformer in which the induced emf per turn of the winding is equal to \(2\ \text{V}\).

Solution.

\[ \Phi_m = \dfrac{e_t}{4.44 f} = \dfrac{2}{4.44(60)} = 0.0075\ \text{Wb} = 7.5\ \text{mWb} \]

Answer: \(7.5\ \text{mWb}\)

2. Equivalent Circuit of an Ideal Transformer

  • Lossless Core: zero hysteresis and zero eddy current losses
  • Zero Winding Resistance: no copper loss in primary or secondary
  • Zero Leakage Flux: total flux links both windings completely

Ideal Transformer Ratios

\[ a = \dfrac{N_1}{N_2} = \dfrac{E_1}{E_2} = \dfrac{V_1}{V_2} \]
\[ \dfrac{I_1}{I_2} = \dfrac{N_2}{N_1} = \dfrac{1}{a} \]
\[ \dfrac{Z_1}{Z_2} = \left(\dfrac{N_1}{N_2}\right)^2 = a^2 \]

Where: \(a\) = turns ratio or transformation ratio, \(N_1, N_2\) = primary and secondary turns, \(E_1, E_2\) = primary and secondary induced voltages (V), \(I_1, I_2\) = primary and secondary currents (A), \(Z_1, Z_2\) = primary and secondary impedances (\(\Omega\))

Example. The low side of a welding transformer has \(2\ \text{turns}\) and delivers \(300\ \text{A}\). If the high voltage side has \(30\ \text{turns}\), how much current flows through it?

Solution.

\[ I_1 = I_2 \left(\dfrac{N_2}{N_1}\right) = 300 \left(\dfrac{2}{30}\right) = 20\ \text{A} \]

Answer: \(20\ \text{A}\)

3. Equivalent Circuit of a Practical Transformer

  • Primary Winding Resistance: resistance of primary coil conductors (\(R_1\))
  • Primary Leakage Reactance: reactance due to primary leakage flux (\(X_1\))
  • Secondary Winding Resistance: resistance of secondary coil conductors (\(R_2\))
  • Secondary Leakage Reactance: reactance due to secondary leakage flux (\(X_2\))
  • Exciting Branch: parallel combination of \(R_m\) and \(X_m\)

Practical Transformer Core Parameters

\[ S_m = E_1 I_0 \qquad\qquad Q_m = \sqrt{S_m^2 - P_m^2} \]
\[ R_m = \dfrac{E_1^2}{P_m} \qquad\qquad X_m = \dfrac{E_1^2}{Q_m} \]

Where: \(S_m\) = core apparent power (VA), \(Q_m\) = magnetizing reactive power (var), \(P_m\) = iron or core loss (W), \(R_m\) = resistance representing iron losses (\(\Omega\)), \(X_m\) = magnetizing reactance (\(\Omega\)), \(E_1\) = primary induced voltage (V), \(I_0\) = exciting current (A)

4. Equivalent Circuit of Practical Transformer at No-Load

  • Open Secondary: secondary current \(I_2 = 0\) at no load
  • Exciting Current: small primary current \(I_0\) flowing into shunt branch
  • Core Loss Component: in-phase current component supplying iron loss (\(I_c\))
  • Magnetizing Component: quadrature current component establishing flux (\(I_m\))

No-Load Exciting Current Relations

\[ I_0 = \sqrt{I_c^2 + I_m^2} \]
\[ I_c = I_0 \cos\theta_0 \qquad\qquad I_m = I_0 \sin\theta_0 \]

Where: \(I_0\) = no-load exciting current (A), \(I_c\) = core loss current component (A), \(I_m\) = magnetizing current component (A), \(\theta_0\) = no-load power factor angle (deg)

5. Equivalent Circuit of Practical Transformer at Full-Load

  • Secondary Load Current: load draws current \(I_2\) at terminal voltage \(V_2\)
  • Primary Load Component: additional primary current \(I_1' = I_2/a\) balancing load
  • Total Primary Current: vector sum of exciting current \(I_0\) and \(I_1'\)

Full-Load Current and Voltage Relations

\[ \mathbf{I}_1 = \mathbf{I}_0 + \mathbf{I}_1' = \mathbf{I}_0 + \dfrac{\mathbf{I}_2}{a} \]
\[ \mathbf{E}_1 = \mathbf{V}_g - \mathbf{I}_1(R_1 + j X_1) \]
\[ \mathbf{V}_2 = \mathbf{E}_2 - \mathbf{I}_2(R_2 + j X_2) \]

Where: \(\mathbf{I}_1\) = total primary line current (A), \(\mathbf{I}_1'\) = load component of primary current (A), \(\mathbf{I}_0\) = exciting current (A), \(\mathbf{V}_g\) = primary terminal supply voltage (V), \(\mathbf{V}_2\) = secondary terminal load voltage (V), \(R_1, R_2\) = primary and secondary resistances (\(\Omega\)), \(X_1, X_2\) = primary and secondary reactances (\(\Omega\))

6. Shifting Transformer Parameters from Primary to Secondary and Vice Versa

  • Referred to Secondary: primary impedance divided by turns ratio squared (\(a^2\))
  • Referred to Primary: secondary impedance multiplied by turns ratio squared (\(a^2\))
  • Equivalent Resistance: total circuit resistance combined on one side
  • Equivalent Reactance: total leakage reactance combined on one side

Parameter Shifting Relations

\[ R_1' = \dfrac{R_1}{a^2} \qquad\qquad X_1' = \dfrac{X_1}{a^2} \qquad\qquad Z_1' = \dfrac{Z_1}{a^2} \]
\[ R_2' = a^2 R_2 \qquad\qquad X_2' = a^2 X_2 \qquad\qquad Z_2' = a^2 Z_2 \]
\[ R_{e1} = R_1 + a^2 R_2 \qquad\qquad X_{e1} = X_1 + a^2 X_2 \]
\[ R_{e2} = R_2 + \dfrac{R_1}{a^2} \qquad\qquad X_{e2} = X_2 + \dfrac{X_1}{a^2} \]

Where: \(R_{e1}, R_{e2}\) = total equivalent resistance referred to primary and secondary (\(\Omega\)), \(X_{e1}, X_{e2}\) = total equivalent reactance referred to primary and secondary (\(\Omega\)), \(a = \dfrac{N_1}{N_2} = \dfrac{V_1}{V_2}\) = turns ratio

7. Equivalent Circuit of a Transformer Referred to the Secondary Side

  • Simplified Model: exciting branch neglected and all series impedance moved to secondary
  • Secondary Series Drop: voltage drop caused by \(I_2 R_{e2}\) and \(I_2 X_{e2}\)
  • Power Angle: angle \(\delta\) between secondary induced voltage \(E_2\) and terminal voltage \(V_2\)

Secondary Referred Circuit Voltage Equation

\[ E_2 = \sqrt{(V_2 \cos\theta + I_2 R_{e2})^2 + (V_2 \sin\theta \pm I_2 X_{e2})^2} \]
\[ \mathbf{E}_2 = V_2 \angle 0^\circ + \mathbf{I}_2 (R_{e2} + j X_{e2}) \]
\[ V_1 = a E_2 \]

Where: \(E_2\) = secondary induced voltage or no-load secondary voltage (V), \(V_2\) = secondary terminal load voltage (V), \(I_2\) = secondary load current (A), \(R_{e2}\) = equivalent resistance referred to secondary (\(\Omega\)), \(X_{e2}\) = equivalent reactance referred to secondary (\(\Omega\)), \(\theta\) = load power factor angle (deg), \(+\) for lagging pf, \(-\) for leading pf

EE Board October 1991

Example. A \(10\ \text{kVA}\), \(2400/240\ \text{V}\) single-phase transformer has \(r_1 = 3\ \Omega, x_1 = 15\ \Omega, r_2 = 0.03\ \Omega, x_2 = 0.15\ \Omega\). Find the primary voltage required to produce \(240\ \text{V}\) at the secondary terminals at full load, \(0.80\) power factor lagging.

Solution.

\[ a = \dfrac{2400}{240} = 10 \implies I_2 = \dfrac{10000}{240} = 41.67\ \text{A} \]
\[ R_{e2} = 0.03 + \dfrac{3}{10^2} = 0.06\ \Omega \qquad\qquad X_{e2} = 0.15 + \dfrac{15}{10^2} = 0.30\ \Omega \]
\[ \mathbf{E}_2 = 240\angle 0^\circ + (41.67\angle -36.87^\circ)(0.06 + j0.30) = 249.65\angle 1.95^\circ\ \text{V} \implies V_1 = 10(249.65) = 2496.5\ \text{V} \]

Answer: \(2496.5\ \text{V}\)

8. Open-Circuit Test of Transformers

  • Low Voltage Side: instruments connected to low voltage winding
  • High Voltage Side Open: high voltage winding left open-circuited
  • Rated Voltage Applied: rated low-side voltage applied during test
  • Core Loss Measurement: wattmeter reading represents constant core/iron losses

Open-Circuit Test Equations

\[ P_{\text{core}} = P_{oc} \]
\[ \cos\theta_0 = \dfrac{P_{oc}}{V_{oc} I_{oc}} \]

Where: \(P_{oc}\) = wattmeter reading during open-circuit test (W), \(P_{\text{core}}\) = total core or iron loss (W), \(V_{oc}\) = applied rated open-circuit voltage (V), \(I_{oc}\) = no-load exciting current (A), \(\cos\theta_0\) = no-load power factor

9. Short-Circuit Test on Transformers

  • High Voltage Side: instruments connected to high voltage winding
  • Low Voltage Side Shorted: low voltage winding short-circuited with thick conductor
  • Reduced Voltage Applied: small voltage applied until rated current flows
  • Copper Loss Measurement: wattmeter reading represents full-load copper loss

Short-Circuit Test Parameters

\[ P_{\text{cuFL}} = P_{sc} \]
\[ Z_e = \dfrac{E_{sc}}{I_{sc}} \qquad\qquad R_e = \dfrac{P_{sc}}{I_{sc}^2} \qquad\qquad X_e = \sqrt{Z_e^2 - R_e^2} \]

Where: \(P_{sc}\) = wattmeter reading during short-circuit test (W), \(P_{\text{cuFL}}\) = full-load copper loss (W), \(E_{sc}\) = applied short-circuit voltage (V), \(I_{sc}\) = rated short-circuit current (A), \(Z_e\) = equivalent impedance (\(\Omega\)), \(R_e\) = equivalent resistance (\(\Omega\)), \(X_e\) = equivalent reactance (\(\Omega\))

EE Board April 1993

Example. A short circuit test was performed upon a \(10\ \text{kVA}\), \(2300/230\ \text{V}\) transformer with results: \(E_{sc} = 137\ \text{V}\), \(P_{sc} = 192\ \text{W}\), \(I_{sc} = 4.34\ \text{A}\). Calculate in primary terms the reactance of the transformer.

Solution.

\[ Z_e = \dfrac{E_{sc}}{I_{sc}} = \dfrac{137}{4.34} = 31.57\ \Omega \qquad\qquad R_e = \dfrac{P_{sc}}{I_{sc}^2} = \dfrac{192}{(4.34)^2} = 10.19\ \Omega \]
\[ X_e = \sqrt{Z_e^2 - R_e^2} = \sqrt{(31.57)^2 - (10.19)^2} = 29.88\ \Omega \]

Answer: \(29.88\ \text{ohms}\)

10. Power Losses and Efficiency

  • Core Loss: sum of eddy current and hysteresis losses (\(P_{\text{core}}\))
  • Copper Loss: \(I^2 R\) loss in primary and secondary windings (\(P_{\text{cu}}\))
  • Efficiency: ratio of output power to input power

Power Losses and Efficiency Equations

\[ P_{\text{in}} = P_{\text{out}} + P_{\text{core}} + P_{\text{cu}} \]
\[ P_{\text{out}} = V_2 I_2 \cos\theta \]
\[ \eta = \dfrac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \dfrac{P_{\text{out}}}{P_{\text{out}} + P_{\text{core}} + P_{\text{cu}}} \times 100\% \]

Where: \(\eta\) = efficiency (\%), \(P_{\text{out}}\) = output power (W), \(P_{\text{in}}\) = input power (W), \(P_{\text{core}}\) = core or iron loss (W), \(P_{\text{cu}}\) = winding copper loss (W), \(V_2\) = load terminal voltage (V), \(I_2\) = load current (A), \(\cos\theta\) = load power factor

EE Board April 1990

Example. A \(50\ \text{kVA}\), \(2300/230\ \text{V}\), \(60\ \text{Hz}\) transformer gives \(187\ \text{W}\) on open-circuit test and \(570\ \text{W}\) on short-circuit test. Determine the efficiency of the transformer at half load, unity power factor.

Solution.

\[ P_{\text{out}} = (0.5)(50000)(1.0) = 25000\ \text{W} \qquad\qquad P_{\text{cu}} = (0.5)^2(570) = 142.5\ \text{W} \]
\[ \eta = \dfrac{25000}{25000 + 187 + 142.5} \times 100\% = 98.70\% \]

Answer: \(98.70\%\)

11. Copper Losses at Any Size of Load

  • Variable Loss: copper loss varies as square of current or load fraction
  • Fractional Load: ratio of operating kVA or current to rated value (\(x\))

Copper Loss at Fractional Load

\[ P_{\text{cu}_x} = x^2 P_{\text{cu}_{\text{FL}}} \]
\[ x = \dfrac{\text{Actual kVA}}{\text{Rated kVA}} = \dfrac{I_{\text{actual}}}{I_{\text{rated}}} \]

Where: \(P_{\text{cu}_x}\) = copper loss at fractional load \(x\) (W), \(P_{\text{cu}_{\text{FL}}}\) = full-load copper loss (W), \(x\) = fractional load or percentage loading

Example. The iron loss of a \(25\ \text{kVA}\), \(60\ \text{Hz}\) single-phase distribution transformer is \(1\%\) of the rated output at unity power factor and full-load copper loss is \(360\ \text{W}\). Determine the efficiency when operating one half of the rated output at \(70\%\) power factor lagging.

Solution.

\[ P_{\text{core}} = 0.01(25000)(1.0) = 250\ \text{W} \qquad\qquad P_{\text{out}} = (0.5)(25000)(0.70) = 8750\ \text{W} \]
\[ P_{\text{cu}} = (0.5)^2(360) = 90\ \text{W} \implies \eta = \dfrac{8750}{8750 + 250 + 90} \times 100\% = 96.26\% \]

Answer: \(96.26\%\)

12. Maximum Efficiency

  • Maximum Condition: variable copper loss equals constant core loss
  • Load at Peak Efficiency: fractional load \(x_m = \sqrt{P_{\text{core}}/P_{\text{cu}_{\text{FL}}}}\)

Maximum Efficiency Equations

\[ x_m = \sqrt{\dfrac{P_{\text{core}}}{P_{\text{cu}_{\text{FL}}}}} \]
\[ \text{kVA}_m = \text{Rated kVA} \times \sqrt{\dfrac{P_{\text{core}}}{P_{\text{cu}_{\text{FL}}}}} \]
\[ \eta_{\max} = \dfrac{P_{\text{out}_m}}{P_{\text{out}_m} + 2 P_{\text{core}}} \times 100\% \]

Where: \(x_m\) = fractional load for maximum efficiency, \(\text{kVA}_m\) = load kVA for maximum efficiency, \(P_{\text{core}}\) = core loss (W), \(P_{\text{cu}_{\text{FL}}}\) = full-load copper loss (W), \(\eta_{\max}\) = maximum efficiency (\%), \(P_{\text{out}_m}\) = output power at maximum efficiency load (W)

Example. A \(10\ \text{kVA}\), \(2300/230\ \text{V}\) transformer has an equivalent resistance referred to the secondary winding of \(0.05\ \Omega\). The core loss is \(75\ \text{W}\). At what kVA load will this transformer operate at maximum efficiency?

Solution.

\[ I_2 = \dfrac{10000}{230} = 43.478\ \text{A} \qquad\qquad P_{\text{cuFL}} = (43.478)^2(0.05) = 94.516\ \text{W} \]
\[ \text{Load} = 10 \times \sqrt{\dfrac{75}{94.516}} = 8.908\ \text{kVA} \approx 8.90\ \text{kVA} \]

Answer: \(8.90\ \text{kVA}\)

13. All-Day Efficiency

  • Energy Efficiency: ratio of 24-hour energy output to 24-hour energy input
  • Distribution Transformer: operates 24 hours continuously with varying load
  • Constant Core Loss Energy: core loss energy equals \(P_{\text{core}} \times 24\ \text{hours}\)

All-Day Efficiency Equations

\[ \eta_{\text{all-day}} = \dfrac{W_{\text{out}}}{W_{\text{out}} + W_{\text{core}} + W_{\text{cu}}} \times 100\% \]
\[ W_{\text{out}} = \sum (\text{Load kVA} \times \text{pf} \times t) \]
\[ W_{\text{core}} = P_{\text{core}} \times 24 \]
\[ W_{\text{cu}} = \sum (x^2 P_{\text{cu}_{\text{FL}}} \times t) \]

Where: \(\eta_{\text{all-day}}\) = all-day energy efficiency (\%), \(W_{\text{out}}\) = total output energy in 24 hours (\(\text{kWh}\)), \(W_{\text{core}}\) = total core loss energy in 24 hours (\(\text{kWh}\)), \(W_{\text{cu}}\) = total copper loss energy in 24 hours (\(\text{kWh}\)), \(t\) = time duration of load (hours)

EE Board October 1980 / EE Board October 1982

Example. A \(25\ \text{kVA}\) distribution transformer operates at \(20\%\) overload, \(0.30\) pf lagging for \(1\ \text{hr}\); full load, \(0.90\) lagging pf for \(3\ \text{hrs}\); \(50\%\) load at unity pf for \(8\ \text{hrs}\); and \(10\%\) load at \(0.80\) leading pf for the rest of the day (\(12\ \text{hrs}\)). If core loss is \(150\ \text{W}\) and full-load copper loss is \(500\ \text{W}\), what is the all-day efficiency?

Solution.

\[ W_{\text{out}} = 25(1.2)(0.3)(1) + 25(1.0)(0.9)(3) + 25(0.5)(1.0)(8) + 25(0.1)(0.8)(12) = 200.5\ \text{kWh} \]
\[ W_{\text{core}} = 0.150(24) = 3.6\ \text{kWh} \qquad\qquad W_{\text{cu}} = 0.500[(1.2)^2(1) + (1.0)^2(3) + (0.5)^2(8) + (0.1)^2(12)] = 3.28\ \text{kWh} \]
\[ \eta_{\text{all-day}} = \dfrac{200.5}{200.5 + 3.6 + 3.28} \times 100\% = 96.68\% \]

Answer: \(96.68\%\)

14. Per-Unit Values

  • Percent Resistance: ratio of full-load resistive drop to rated voltage
  • Percent Reactance: ratio of full-load reactive drop to rated voltage
  • Percent Impedance: ratio of short-circuit voltage to rated voltage

Percent and Per-Unit Parameter Equations

\[ \%\text{IR} = \dfrac{I_2 R_{e2}}{V_2} \times 100\% = \dfrac{P_{sc}}{\text{VA}_{\text{rating}}} \times 100\% \]
\[ \%\text{IZ} = \dfrac{E_{sc}}{V_{\text{rated}}} \times 100\% = \dfrac{I_2 Z_{e2}}{V_2} \times 100\% \]
\[ \%\text{IX} = \sqrt{(\%\text{IZ})^2 - (\%\text{IR})^2} \]

Where: \(\%\text{IR}\) = percent resistance, \(\%\text{IX}\) = percent reactance, \(\%\text{IZ}\) = percent impedance, \(P_{sc}\) = short-circuit test copper loss (W), \(E_{sc}\) = short-circuit test voltage (V), \(\text{VA}_{\text{rating}}\) = rated apparent power (VA), \(V_{\text{rated}}\) = rated voltage (V)

EE Board April 1983

Example. A single-phase transformer rated \(75\ \text{kVA}\), \(8000\ \text{V}\) primary \(240\ \text{V}\) secondary, is given the short circuit test. With the secondary terminals short-circuited, \(440\ \text{V}\) (\(60\ \text{Hz}\)) is impressed on the primary, which then draws rated current and \(1125\ \text{W}\). Solve for the percentage impedance of the transformer.

Solution.

\[ \%\text{IZ} = \dfrac{E_{sc}}{V_{\text{rated}}} \times 100\% = \dfrac{440}{8000} \times 100\% = 5.50\% \]

Answer: \(5.50\%\)

15. Voltage Regulation

  • Definition: percentage change in secondary terminal voltage from full load to no load
  • Lagging Load: voltage drops under load, causing positive regulation
  • Leading Load: voltage rise possible under load, causing negative regulation

Voltage Regulation Equations

\[ \%\text{VR} = \dfrac{E_2 - V_2}{V_2} \times 100\% \]
\[ \%\text{VR} = \sqrt{(\cos\theta + \%\text{IR})^2 + (\sin\theta \pm \%\text{IX})^2} - 1 \]

Where: \(\%\text{VR}\) = percent voltage regulation, \(E_2\) = secondary no-load voltage (V), \(V_2\) = secondary full-load voltage (V), \(\%\text{IR}\) = percent resistance, \(\%\text{IX}\) = percent reactance, \(\theta\) = load power factor angle (deg), \(+\) for lagging pf, \(-\) for leading pf

EE Board April 1990

Example. Short circuit test on a \(100\ \text{kVA}\), \(2400/240\ \text{V}\) distribution transformer gave: \(E_{sc} = 72\ \text{V}\), \(I_{sc} = 41.6\ \text{A}\), \(P_{sc} = 1180\ \text{W}\). Calculate the percent regulation at a power factor of \(0.75\) lagging.

Solution.

\[ \%\text{IZ} = \dfrac{72}{2400} \times 100\% = 3.0\% \qquad\qquad \%\text{IR} = \dfrac{1180}{100000} \times 100\% = 1.18\% \]
\[ \%\text{IX} = \sqrt{(3.0)^2 - (1.18)^2} = 2.758\% \]
\[ \%\text{VR} = \sqrt{(0.75 + 0.0118)^2 + (0.6614 + 0.02758)^2} - 1 = 0.0271 = 2.71\% \]

Answer: \(2.71\%\)

16. Autotransformer

  • Single Winding: one continuous winding common to both primary and secondary
  • Common Winding Current: difference between primary and secondary currents (\(I_c\))
  • Transformed Power: power transferred via magnetic induction
  • Conducted Power: power transferred directly via electrical conduction

Autotransformer Equations

\[ a = \dfrac{V_1}{V_2} = \dfrac{N_1}{N_2} = \dfrac{I_2}{I_1} \]
\[ I_c = |I_1 - I_2| \]
\[ P_{\text{transformed}} = P_{\text{input}} \left(1 - \dfrac{1}{a}\right) \]
\[ P_{\text{conducted}} = \dfrac{P_{\text{input}}}{a} \]

Where: \(a\) = turns or voltage ratio (\(V_1/V_2 > 1\) for step-down), \(V_1, V_2\) = primary and secondary voltages (V), \(I_1, I_2\) = primary and secondary currents (A), \(I_c\) = common winding current (A), \(P_{\text{input}}\) = total input VA or power (W)

Example. A \(20000\ \text{VA}\), \(400/500\ \text{V}\) autotransformer operates at full load. Determine the current flowing in the common winding.

Solution.

\[ I_1 = \dfrac{20000}{400} = 50\ \text{A} \qquad\qquad I_2 = \dfrac{20000}{500} = 40\ \text{A} \]
\[ I_c = I_1 - I_2 = 50 - 40 = 10\ \text{A} \]

Answer: \(10\ \text{A}\)

17. Current Transformer

  • Series Instrument Transformer: primary connected in series with high current line
  • Ammeter Secondary: secondary connected to standard 5-ampere AC ammeter
  • Line Current Formula: line current equals ammeter reading times CT ratio

Current Transformer Equation

\[ I_L = A \times \text{CT ratio} \]
\[ \text{CT ratio} = \dfrac{I_{\text{primary}}}{I_{\text{secondary}}} \]

Where: \(I_L\) = actual line current to be measured (A), \(A\) = ammeter reading (A), \(\text{CT ratio}\) = nominal current transformation ratio of current transformer

Example. An ammeter connected to a \(100/5\) current transformer reads \(4\ \text{A}\). Calculate the actual line current.

Solution.

\[ I_L = A \times \text{CT ratio} = 4 \times \left(\dfrac{100}{5}\right) = 80\ \text{A} \]

Answer: \(80\ \text{A}\)

18. Potential Transformer

  • Parallel Instrument Transformer: primary connected across high voltage line
  • Voltmeter Secondary: secondary connected to standard 115-volt AC voltmeter
  • Line Voltage Formula: line voltage equals voltmeter reading times PT ratio

Potential Transformer Equation

\[ V_L = V \times \text{PT ratio} \]
\[ \text{PT ratio} = \dfrac{V_{\text{primary}}}{V_{\text{secondary}}} \]

Where: \(V_L\) = actual high voltage line potential (V), \(V\) = voltmeter reading (V), \(\text{PT ratio}\) = nominal voltage transformation ratio of potential transformer

EE Board October 1998

Example. The potential transformer of a line has a ratio of \(132000/66\). The voltmeter reads \(64\ \text{V}\). What is the voltage of the line?

Solution.

\[ V_L = V \times \text{PT ratio} = 64 \times \left(\dfrac{132000}{66}\right) = 128000\ \text{V} = 128\ \text{kV} \]

Answer: \(128\ \text{kV}\)

15. Formula Summary

Formula Name Formula Variables
Induced EMF Equation \(E = 4.44 N f \Phi_m\) \(E\) = induced rms voltage (V)
\(N\) = number of turns
\(f\) = frequency (Hz)
\(\Phi_m\) = peak flux (Wb)
Induced EMF per Turn \(e_t = 4.44 f \Phi_m\) \(e_t\) = induced emf per turn (V/turn)
\(f\) = frequency (Hz)
\(\Phi_m\) = peak flux (Wb)
Peak Magnetic Flux \(\Phi_m = \beta_m A\) \(\Phi_m\) = peak flux (Wb)
\(\beta_m\) = flux density (T)
\(A\) = core area (\(\text{m}^2\))
Ideal Transformer Voltage Ratio \(a = \dfrac{N_1}{N_2} = \dfrac{E_1}{E_2} = \dfrac{V_1}{V_2}\) \(a\) = turns ratio
\(N_1, N_2\) = primary and secondary turns
\(V_1, V_2\) = primary and secondary voltages (V)
Ideal Transformer Current Ratio \(\dfrac{I_1}{I_2} = \dfrac{N_2}{N_1} = \dfrac{1}{a}\) \(I_1, I_2\) = primary and secondary currents (A)
\(a\) = turns ratio
Ideal Transformer Impedance Ratio \(\dfrac{Z_1}{Z_2} = \left(\dfrac{N_1}{N_2}\right)^2 = a^2\) \(Z_1, Z_2\) = primary and secondary impedances (\(\Omega\))
\(a\) = turns ratio
Core Loss Resistance \(R_m = \dfrac{E_1^2}{P_m}\) \(R_m\) = core loss resistance (\(\Omega\))
\(E_1\) = primary induced voltage (V)
\(P_m\) = core loss (W)
Magnetizing Reactance \(X_m = \dfrac{E_1^2}{Q_m}\) \(X_m\) = magnetizing reactance (\(\Omega\))
\(E_1\) = primary induced voltage (V)
\(Q_m\) = reactive magnetizing power (var)
No-Load Exciting Current \(I_0 = \sqrt{I_c^2 + I_m^2}\) \(I_0\) = exciting current (A)
\(I_c\) = core loss current component (A)
\(I_m\) = magnetizing current component (A)
Exciting Current Components \(I_c = I_0 \cos\theta_0\)
\(I_m = I_0 \sin\theta_0\)		 \(I_c\) = core loss current (A)
\(I_m\) = magnetizing current (A)
\(\theta_0\) = no-load power factor angle (deg)
Total Primary Current \(\mathbf{I}_1 = \mathbf{I}_0 + \mathbf{I}_1' = \mathbf{I}_0 + \dfrac{\mathbf{I}_2}{a}\) \(\mathbf{I}_1\) = total primary current (A)
\(\mathbf{I}_0\) = exciting current (A)
\(\mathbf{I}_2\) = secondary current (A)
\(a\) = turns ratio
Equivalent Primary Resistance \(R_{e1} = R_1 + a^2 R_2\) \(R_{e1}\) = total resistance referred to primary (\(\Omega\))
\(R_1, R_2\) = primary and secondary resistances (\(\Omega\))
\(a\) = turns ratio
Equivalent Primary Reactance \(X_{e1} = X_1 + a^2 X_2\) \(X_{e1}\) = total reactance referred to primary (\(\Omega\))
\(X_1, X_2\) = primary and secondary reactances (\(\Omega\))
\(a\) = turns ratio
Equivalent Secondary Resistance \(R_{e2} = R_2 + \dfrac{R_1}{a^2}\) \(R_{e2}\) = total resistance referred to secondary (\(\Omega\))
\(R_1, R_2\) = primary and secondary resistances (\(\Omega\))
\(a\) = turns ratio
Equivalent Secondary Reactance \(X_{e2} = X_2 + \dfrac{X_1}{a^2}\) \(X_{e2}\) = total reactance referred to secondary (\(\Omega\))
\(X_1, X_2\) = primary and secondary reactances (\(\Omega\))
\(a\) = turns ratio
Secondary Induced Voltage \(E_2 = \sqrt{(V_2 \cos\theta + I_2 R_{e2})^2 + (V_2 \sin\theta \pm I_2 X_{e2})^2}\) \(E_2\) = secondary no-load voltage (V)
\(V_2\) = secondary load voltage (V)
\(I_2\) = secondary current (A)
\(R_{e2}\) = equivalent resistance (\(\Omega\))
\(X_{e2}\) = equivalent reactance (\(\Omega\))
Open-Circuit Power Factor \(\cos\theta_0 = \dfrac{P_{oc}}{V_{oc} I_{oc}}\) \(\cos\theta_0\) = no-load power factor
\(P_{oc}\) = open-circuit test power (W)
\(V_{oc}\) = open-circuit voltage (V)
\(I_{oc}\) = no-load current (A)
Equivalent Short-Circuit Impedance \(Z_e = \dfrac{E_{sc}}{I_{sc}}\) \(Z_e\) = equivalent impedance (\(\Omega\))
\(E_{sc}\) = short-circuit test voltage (V)
\(I_{sc}\) = rated short-circuit current (A)
Equivalent Short-Circuit Resistance \(R_e = \dfrac{P_{sc}}{I_{sc}^2}\) \(R_e\) = equivalent resistance (\(\Omega\))
\(P_{sc}\) = short-circuit test power (W)
\(I_{sc}\) = short-circuit current (A)
Equivalent Short-Circuit Reactance \(X_e = \sqrt{Z_e^2 - R_e^2}\) \(X_e\) = equivalent reactance (\(\Omega\))
\(Z_e\) = equivalent impedance (\(\Omega\))
\(R_e\) = equivalent resistance (\(\Omega\))
Transformer Efficiency \(\eta = \dfrac{P_{\text{out}}}{P_{\text{out}} + P_{\text{core}} + P_{\text{cu}}} \times 100\%\) \(\eta\) = efficiency (\%)
\(P_{\text{out}}\) = output power (W)
\(P_{\text{core}}\) = core loss (W)
\(P_{\text{cu}}\) = copper loss (W)
Copper Loss at Fractional Load \(P_{\text{cu}_x} = x^2 P_{\text{cu}_{\text{FL}}}\) \(P_{\text{cu}_x}\) = copper loss at fractional load \(x\) (W)
\(x\) = load fraction
\(P_{\text{cu}_{\text{FL}}}\) = full-load copper loss (W)
Fractional Load for Max Efficiency \(x_m = \sqrt{\dfrac{P_{\text{core}}}{P_{\text{cu}_{\text{FL}}}}}\) \(x_m\) = load fraction at maximum efficiency
\(P_{\text{core}}\) = core loss (W)
\(P_{\text{cu}_{\text{FL}}}\) = full-load copper loss (W)
Maximum Efficiency Load kVA \(\text{kVA}_m = \text{Rated kVA} \times \sqrt{\dfrac{P_{\text{core}}}{P_{\text{cu}_{\text{FL}}}}}\) \(\text{kVA}_m\) = load kVA at max efficiency
\(P_{\text{core}}\) = core loss (W)
\(P_{\text{cu}_{\text{FL}}}\) = full-load copper loss (W)
All-Day Efficiency \(\eta_{\text{all-day}} = \dfrac{W_{\text{out}}}{W_{\text{out}} + W_{\text{core}} + W_{\text{cu}}} \times 100\%\) \(\eta_{\text{all-day}}\) = all-day efficiency (\%)
\(W_{\text{out}}\) = total output energy (\(\text{kWh}\))
\(W_{\text{core}}\) = core loss energy (\(\text{kWh}\))
\(W_{\text{cu}}\) = copper loss energy (\(\text{kWh}\))
Percent Resistance \(\%\text{IR} = \dfrac{I_2 R_{e2}}{V_2} \times 100\% = \dfrac{P_{sc}}{\text{VA}_{\text{rating}}} \times 100\%\) \(\%\text{IR}\) = percent resistance
\(I_2 R_{e2}\) = resistive drop (V)
\(P_{sc}\) = short-circuit power (W)
Percent Impedance \(\%\text{IZ} = \dfrac{E_{sc}}{V_{\text{rated}}} \times 100\% = \dfrac{I_2 Z_{e2}}{V_2} \times 100\%\) \(\%\text{IZ}\) = percent impedance
\(E_{sc}\) = short-circuit voltage (V)
\(V_{\text{rated}}\) = rated voltage (V)
Percent Reactance \(\%\text{IX} = \sqrt{(\%\text{IZ})^2 - (\%\text{IR})^2}\) \(\%\text{IX}\) = percent reactance
\(\%\text{IZ}\) = percent impedance
\(\%\text{IR}\) = percent resistance
Percent Voltage Regulation \(\%\text{VR} = \dfrac{E_2 - V_2}{V_2} \times 100\%\) \(\%\text{VR}\) = percent voltage regulation
\(E_2\) = no-load voltage (V)
\(V_2\) = full-load voltage (V)
Voltage Regulation from Percent Drops \(\%\text{VR} = \sqrt{(\cos\theta + \%\text{IR})^2 + (\sin\theta \pm \%\text{IX})^2} - 1\) \(\%\text{VR}\) = percent voltage regulation
\(\theta\) = power factor angle (deg)
\(\%\text{IR}\) = percent resistance
\(\%\text{IX}\) = percent reactance
Autotransformer Common Winding Current \(I_c = |I_1 - I_2|\) \(I_c\) = common winding current (A)
\(I_1, I_2\) = primary and secondary currents (A)
Autotransformer Transformed Power \(P_{\text{transformed}} = P_{\text{input}} \left(1 - \dfrac{1}{a}\right)\) \(P_{\text{transformed}}\) = power transferred by induction (W)
\(P_{\text{input}}\) = total input power (W)
\(a\) = voltage ratio (\(V_1/V_2 > 1\))
Current Transformer Line Current \(I_L = A \times \text{CT ratio}\) \(I_L\) = line current (A)
\(A\) = ammeter reading (A)
\(\text{CT ratio}\) = primary to secondary current ratio
Potential Transformer Line Voltage \(V_L = V \times \text{PT ratio}\) \(V_L\) = line voltage (V)
\(V\) = voltmeter reading (V)
\(\text{PT ratio}\) = primary to secondary voltage ratio