The short circuit test stands as one of the most practical methods to evaluate transformer performance without requiring actual load conditions. This test reveals critical parameters that help engineers predict how a transformer will behave under real operating conditions, particularly its copper losses and internal impedance characteristics.
What is Short Circuit Test?
A short circuit test involves deliberately connecting the secondary winding terminals with a thick conductor while applying a carefully controlled low voltage to the primary winding. The voltage applied is gradually increased until the transformer draws its rated full-load current. Despite the name, this test is perfectly safe when performed correctly, as the applied voltage rarely exceeds 5-10% of the normal rated voltage.
The fundamental purpose of conducting this test goes beyond simple measurement. Engineers use it to determine the full-load copper losses in transformer windings, calculate equivalent resistance and reactance values, and establish the transformer’s percentage impedance. These parameters become essential when predicting voltage regulation and calculating efficiency under various load conditions.
Why Perform on High Voltage Side?
An interesting aspect of the short circuit test is that it’s typically performed on the high voltage side while shorting the low voltage terminals. This practice stems from practical considerations rather than theoretical requirements.
The high voltage winding carries lower current compared to the low voltage side at the same power rating. Conducting the test from the HV side means using instruments with lower current ranges, which are more economical and readily available in testing laboratories. Additionally, the reduced current requirement translates to lower power consumption during testing, making the entire process more efficient and safer.
The thick copper conductor used to short the LV terminals must have sufficient cross-sectional area to handle the full-load current without overheating. A current density of approximately 2.5 A/mm² is typically recommended, and the connection should be as short as possible with excellent contact points to minimize additional resistance.
Circuit Arrangement and Equipment
The test setup requires specific instruments connected in a particular configuration. On the primary (HV) side, connect a wattmeter to measure input power, an ammeter in series to monitor current flow, and a voltmeter to measure the applied voltage. The secondary (LV) side gets short-circuited using a thick copper bar or conductor.
A variable voltage source, typically an autotransformer or variac, supplies the controlled voltage to the primary winding. This arrangement allows for gradual voltage increase from zero to the level needed to circulate rated current.
Test Procedure Step by Step
Before starting, verify that all instruments are properly calibrated and the transformer is at ambient temperature. The winding under test should be set to its rated tap position.
Begin with the variac set to its minimum position, producing zero output voltage. Short circuit the low voltage terminals using an appropriate thick conductor, ensuring firm connections. Switch on the power supply and gradually increase the primary voltage while carefully monitoring the ammeter reading.
Continue raising the voltage slowly until the ammeter indicates the transformer’s rated full-load current. At this point, immediately record all three meter readings: the voltmeter showing applied voltage, the ammeter displaying current, and the wattmeter indicating power consumption. Read the meters quickly to prevent winding temperature rise.
After recording the readings, reduce the voltage back to zero before switching off the power supply. The entire test typically takes only a few minutes, and the transformer barely heats up due to the low voltage applied.
Understanding the Measurements
The instruments provide three crucial readings that unlock the transformer’s internal characteristics. The voltmeter reading represents the voltage required to overcome the transformer’s internal impedance and circulate full-load current. This voltage, expressed as a percentage of the rated voltage, gives the transformer’s percentage impedance – a key parameter for fault current calculations.
The wattmeter reading during this test has special significance. Since the applied voltage is so low (only 5-10% of rated voltage), the flux in the transformer core is proportionally reduced. Core losses depend on the magnetic flux density, so at such low flux levels, iron losses become negligibly small. The wattmeter therefore reads almost exclusively the copper losses occurring in both primary and secondary windings at full-load current.
The ammeter confirms that rated current is flowing through both windings. This ensures that the copper losses measured represent actual full-load conditions, making the test results directly applicable to real operating scenarios.
Mathematical Calculations
The recorded readings enable calculation of several important parameters using straightforward formulas. Let’s denote the short circuit test readings as \(V_{sc}\) for voltage, \(I_{sc}\) for current, and \(W_{sc}\) for power.
The equivalent resistance referred to the primary side is calculated from the relationship between power dissipation and current:
\(R_{01}=\frac{W_{sc}}{I^2_{sc}}\)
This resistance represents the combined effect of primary and secondary winding resistances as seen from the primary side.
The equivalent impedance comes from the ratio of applied voltage to current:
\(Z_{01}=\frac{V_{sc}}{I_{sc}}\)
Using these two values, the equivalent reactance can be determined:
\(X_{01}=\sqrt{Z^2_{01}-{R^2_{01}}}\)
The percentage impedance, crucial for protective relay coordination and fault calculations, is simply:
\(\%Z=\frac{V_{sc}}{V_{rated}}\times 100\)
This tells what percentage of rated voltage is needed to drive full-load current through the transformer’s internal impedance.
Short Circuit Test Voltage vs Current Graph
📈 Voltage vs Current Graph
Short Circuit Test Interactive Visualization
Practical Example
Consider a transformer where the short circuit test yielded the following readings: applied voltage of \(15 V\), short circuit current of \(10 A\), and wattmeter reading of \(120 W\). If the transformer's rated primary voltage is \(240 V\), we can calculate all parameters.
The equivalent resistance works out to \(1.20 Ω\), found by dividing \(120 W\) by the square of \(10 A\). The equivalent impedance equals \(1.50 Ω\), obtained by dividing \(15 V\) by \(10 A\). The equivalent reactance then calculates to \(0.90 Ω\) using the impedance triangle relationship. The percentage impedance comes to \(6.25\%\), which is the ratio of \(15 V\) to \(240 V\) multiplied by \(100\). Finally, the power factor during short circuit is \(0.800\), calculated from the ratio of real power to apparent power.
These values provide complete information about the transformer's series impedance characteristics and full-load copper losses, enabling accurate prediction of voltage regulation and efficiency calculations.
Problem 1: During a short circuit test on a single-phase transformer, the following readings were obtained:
- Short circuit voltage, \(V_{sc}\) = 12.5 V
- Short circuit current, \(I_{sc}\) = 8.0 A
- Wattmeter reading, \(W_{sc}\) = 85 W
- Rated primary voltage, \(V_{rated}\) = 230 V
Calculate:
- (a) Equivalent resistance referred to primary \((R_{01})\)
- (b) Equivalent impedance referred to primary \((Z_{01})\)
- (c) Equivalent reactance referred to primary \((X_{01})\)
- (d) Percentage impedance \((\%Z)\)
- (e) Power factor during short circuit test \((\cos\phi)\)
Solution:
(a) Equivalent Resistance:
\(R_{01} = \frac{W_{sc}}{I_{sc}^2} = \frac{85}{8.0^2} = \frac{85}{64}\)
\(R_{01} = 1.3281 \) Ω
(b) Equivalent Impedance:
\(Z_{01} = \frac{V_{sc}}{I_{sc}} = \frac{12.5}{8.0}\)
\(Z_{01} = 1.5625\) Ω
(c) Equivalent Reactance:
\(X_{01} = \sqrt{Z_{01}^2 - R_{01}^2} = \sqrt{(1.5625)^2 - (1.3281)^2}\)
\(X_{01} = \sqrt{2.4414 - 1.7639} = \sqrt{0.6775}\)
\(X_{01} = 0.8231\) Ω
(d) Percentage Impedance:
\(\%Z = \frac{V_{sc}}{V_{rated}} \times 100 = \frac{12.5}{230} \times 100\)
\(\%Z = 5.43\%\)
(e) Power Factor:
\(\cos\phi = \frac{W_{sc}}{V_{sc} \times I_{sc}} = \frac{85}{12.5 \times 8.0} = \frac{85}{100}\)
\(\cos\phi = 0.8500\)
Problem 2: A short circuit test on a transformer gave the following data:
\(V_{sc}\) = 18 V, \(I_{sc}\) = 12 A, \(W_{sc}\) = 150 W, \(V_{rated}\) = 400 V
Determine the copper losses when the transformer operates at:
- (a) Half load (50%)
- (b) Three-quarter load (75%)
Solution:
The full-load copper loss from the test is \(W_{sc}\) = 150 W
Copper loss varies with the square of the load current:
\(P_{cu(x)} = P_{cu(FL)} \times \left(\frac{x}{100}\right)^2\)
(a) Copper loss at half load (50%):
\(P_{cu(0.5)} = 150 \times (0.5)^2 = 150 \times 0.25\)
\(P_{cu(0.5)} = 37.50\) W
(b) Copper loss at three-quarter load (75%):
\(P_{cu(0.75)} = 150 \times (0.75)^2 = 150 \times 0.5625\)
\(P_{cu(0.75)} = 84.38\) W
Problem 3: In a short circuit test on a transformer, the following observations were made:
\(V_{sc}\) = 20 V, \(I_{sc}\) = 15 A, \(W_{sc}\) = 200 W
The percentage impedance of the transformer is 5%. Calculate:
- (a) Rated primary voltage \((V_{rated})\)
- (b) Equivalent resistance \((R_{01})\)
- (c) Equivalent impedance \((Z_{01})\)
- (d) Equivalent reactance \((X_{01})\)
Solution:
(a) Rated Primary Voltage:
From percentage impedance formula:
\(\%Z = \frac{V_{sc}}{V_{rated}} \times 100\)
\(V_{rated} = \frac{V_{sc} \times 100}{\%Z} = \frac{20 \times 100}{5}\)
\(V_{rated} = 400\) V
(b) Equivalent Resistance:
\(R_{01} = \frac{W_{sc}}{I_{sc}^2} = \frac{200}{15^2} = \frac{200}{225}\)
\(R_{01} = 0.8889\) Ω
(c) Equivalent Impedance:
\(Z_{01} = \frac{V_{sc}}{I_{sc}} = \frac{20}{15}\)
\(Z_{01} = 1.3333\) Ω
(d) Equivalent Reactance:
\(X_{01} = \sqrt{Z_{01}^2 - R_{01}^2} = \sqrt{(1.3333)^2 - (0.8889)^2}\)
\(X_{01} = 0.9938\) Ω
Problem 4: A transformer has the following equivalent circuit parameters referred to primary:
\(R_{01}\) = 1.2 Ω, \(X_{01}\) = 0.9 Ω
The transformer has a rated current of 10 A and rated primary voltage of 240 V.
Determine the readings that would be obtained during a short circuit test:
- (a) Equivalent impedance \((Z_{01})\)
- (b) Short circuit voltage \((V_{sc})\)
- (c) Wattmeter reading \((W_{sc})\)
- (d) Percentage impedance \((\%Z)\)
Solution:
(a) Equivalent Impedance:
\(Z_{01} = \sqrt{R_{01}^2 + X_{01}^2} = \sqrt{(1.2)^2 + (0.9)^2}\)
\(Z_{01} = \sqrt{1.44 + 0.81} = \sqrt{2.25}\)
\(Z_{01} = 1.50\) Ω
(b) Short Circuit Voltage:
At rated current:
\(V_{sc} = Z_{01} \times I_{rated} = 1.5 \times 10\)
\(V_{sc} = 15.0\) V
(c) Wattmeter Reading (Full-load copper loss):
\(W_{sc} = R_{01} \times I_{rated}^2 = 1.2 \times 10^2 = 1.2 \times 100\)
\(W_{sc} = 120.0\) W
(d) Percentage Impedance:
\(\%Z = \frac{V_{sc}}{V_{rated}} \times 100 = \frac{15}{240} \times 100\)
\(\%Z = 6.25\%\)
Problem 5: A short circuit test was performed on a 440 V transformer and the results obtained were:
\(V_{sc}\) = 22 V, \(I_{sc}\) = 20 A, \(W_{sc}\) = 320 W
The transformer supplies a load operating at 80% of full load with power factor 0.8 lagging.
Calculate:
- (a) Equivalent resistance, impedance, and reactance
- (b) Percentage impedance
- (c) Copper loss at 80% load
- (d) Voltage drop in the transformer at this load
Solution:
(a) Equivalent Parameters:
\(R_{01} = \frac{W_{sc}}{I_{sc}^2} = \frac{320}{20^2} = \frac{320}{400}\)
\(R_{01} = 0.80\) Ω
\(Z_{01} = \frac{V_{sc}}{I_{sc}} = \frac{22}{20}\)
\(Z_{01} = 1.10\) Ω
\(X_{01} = \sqrt{Z_{01}^2 - R_{01}^2} = \sqrt{(1.1)^2 - (0.8)^2} = \sqrt{1.21 - 0.64}\)
\(X_{01} = 0.7550$\) Ω
(b) Percentage Impedance:
\(\%Z = \frac{V_{sc}}{V_{rated}} \times 100 = \frac{22}{440} \times 100\)
\(\%Z = 5\%\)
(c) Copper Loss at 80% Load:
\(P_{cu(0.8)} = P_{cu(FL)} \times (0.8)^2 = 320 \times 0.64\)
\(P_{cu(0.8)} = 204.80\) W
(d) Voltage Drop:
Load current: \(I_{load} = 0.8 \times 20 = 16\) A
Power factor: \(\cos\phi = 0.8\), so \(\sin\phi = 0.6\)
\(V_{drop} = I_{load}(R_{01}\cos\phi + X_{01}\sin\phi)\)
\(V_{drop} = 16(0.8 \times 0.8 + 0.7550 \times 0.6)\)
\(V_{drop} = 16(0.64 + 0.453)\)
\(V_{drop} = 17.49\) V
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Applications and Importance
The short circuit test delivers information that proves invaluable across multiple engineering applications. Voltage regulation calculations rely heavily on knowing the equivalent resistance and reactance values. These parameters allow engineers to predict how much the secondary voltage will drop when load is applied, ensuring the transformer meets performance specifications.
Efficiency calculations require accurate knowledge of copper losses at various load levels. Since copper losses vary with the square of the load current, the full-load copper loss measured in this test becomes the reference for calculating losses at any other load condition.
Transformer impedance affects fault current levels throughout the electrical system. Protection engineers use the percentage impedance determined from this test to set protective relay parameters and ensure adequate short circuit interrupting capacity in circuit breakers.
The test also serves quality control purposes during transformer manufacturing. Comparing test results against design specifications helps identify manufacturing defects such as incorrect turns, poor connections, or winding faults before the transformer leaves the factory.
Safety Precautions
While the short circuit test uses low voltage, several safety measures remain essential. Always verify that the transformer is properly grounded before beginning the test. The test area should be clearly marked and restricted to authorized personnel only.
Use test leads with adequate cross-sectional area to handle the full-load current without overheating. Poor connections can create additional resistance that affects measurement accuracy and poses a fire hazard.
Never attempt to change connections while voltage is applied. Always reduce the variac to zero and switch off the power supply before making any modifications to the circuit.
Monitor the ammeter continuously during voltage increase to avoid exceeding rated current. Operating beyond rated current, even briefly, can overheat the windings and damage the transformer.
Wear appropriate personal protective equipment including insulated gloves and safety glasses. Even though the voltage is low during this test, the high currents involved present their own hazards.
Difference from Open Circuit Test
Understanding both the short circuit and open circuit tests provides a complete picture of transformer characteristics. While the short circuit test measures copper losses and series parameters, the open circuit test determines core losses and shunt parameters.
The open circuit test is performed on the low voltage side with the high voltage side kept open, exactly opposite to the short circuit test arrangement. It uses rated voltage applied to the LV winding while the HV side carries no load.
The open circuit test measures core losses that remain constant regardless of load, while the short circuit test measures copper losses that vary with load current. Together, these two tests provide all the information needed to construct the complete equivalent circuit of the transformer without requiring actual load testing.
Conclusion
The short circuit test represents an economical and accurate method for determining critical transformer parameters. By applying just 5-10% of rated voltage, this test safely measures full-load copper losses, equivalent impedance, and other characteristics that would otherwise require loading the transformer to its full capacity.
The simplicity of the test procedure combined with the wealth of information it provides makes it indispensable in transformer testing protocols. Whether in manufacturing facilities, testing laboratories, or field installations, this test continues to serve as a fundamental tool for evaluating transformer performance and ensuring reliable operation in power systems.
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