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Short-circuit and open-circuit tests are to be performed on a 3-winding transformer to determine the parameters of the equivalent circuits of it.

The equivalent impedances Z_{1}, Z_{2} and Z_{3} referred to a common base can be determined by performing three short-circuit tests as follows −

In the first test, the secondary winding is short circuited, the tertiary winding is kept open circuited and a low voltage being applied to the primary winding so that full-load current flows in the secondary winding (see the figure above). Then, the voltage, current and power input to the primary winding is to be measured.

Let V_{1}, I_{1} and P_{1} be the voltmeter, ammeter and wattmeter readings respectively. Now, if Z_{12} is the short circuit impedance of primary and secondary windings with tertiary winding open. Then,

$$\mathrm{Equivalent\:impedance,\:𝑍_{12} =\frac{𝑉_{1}}{𝐼_{1}}}$$

$$\mathrm{Equivalent\:resistance,\:𝑅_{12} =\frac{𝑃_{1}}{{𝐼^2}_{1}}}$$

And,

$$\mathrm{Equivalent\:leakage\:reactance,\:𝑋_{12} =\sqrt{{𝑍^2}_{12} − {𝑅^2}_{12}}}$$

From the equivalent circuit it can be seen that the impedance Z_{12} is a series combination of Z_{1} and Z_{2}, i.e.,

$$\mathrm{𝑍_{12} = 𝑅_{12} + 𝑗𝑋_{12} = 𝑍_{1} + 𝑍_{2} … (1)}$$

In the second test, the tertiary winding is short circuited and the secondary winding is kept
open. A low voltage is applied to the primary winding, which circulates full-load current in the
tertiary winding. If Z_{13} is the short-circuit impedance of primary and tertiary windings with secondary winding open, then,

$$\mathrm{𝑍_{13} = 𝑍_{1} + 𝑍_{3} … (2)}$$

In the third test, the tertiary winding is short circuited and the primary winding is kept open. A low voltage is applied to the secondary winding so that full-load current flows in the short circuited tertiary winding. If Z_{23} is the short circuit impedance of secondary and tertiary windings with primary open, then,

$$\mathrm{𝑍_{23} = 𝑍_{2} + 𝑍_{3} … (3)}$$

Now, from the eqns. (1), (2) and (3), the impedances Z_{1}, Z_{2} and Z_{3} can be obtained (all should be referred to primary side) as follows −

$$\mathrm{𝑍_{1} =\frac{1}{2}(𝑍_{12} + 𝑍_{13} − 𝑍_{23}) … (4)}$$

$$\mathrm{𝑍_{2} =\frac{1}{2}(𝑍_{23} + 𝑍_{12} − 𝑍_{13}) … (5)}$$

$$\mathrm{𝑍_{3} =\frac{1}{2}(𝑍_{13} + 𝑍_{23} − 𝑍_{12}) … (6)}$$

Where,

$$\mathrm{𝑍_{1} = 𝑅_{1} + 𝑗𝑋_{1}; 𝑍_{2} = 𝑅_{2} + 𝑗𝑋_{2}; 𝑍_{3} = 𝑅_{3} + 𝑗𝑋_{3}}$$

**Note** – The impedances Z_{12} and Z_{13} are referred to primary side since the instruments are connected in the primary side. The impedance Z_{23} is referred to secondary side. Hence, it has to be referred to the primary side and only then eqns. (4), (5) & (6) can be used for calculating Z_{1}, Z_{2} and Z_{3}.

The open circuit test on a three winding transformer is performed to determine the core loss, magnetising impedance and turn ratios. In the open circuit test, the primary winding is energized with both secondary and tertiary windings on open circuit. Then, we get,

$$\mathrm{Voltage\:ratio\:between\:primary\:and \:secondary ,\:𝐾_{12} =\frac{𝑉_{1}}{𝑉_{2}}}$$

$$\mathrm{Voltage\:ratio\:between\:primary\:and \:tertiary ,\:𝐾_{13} =\frac{𝑉_{1}}{𝑉_{3}}}$$

$$\mathrm{Voltage\:ratio\:between\:secondary \:and\:tertiary ,\:𝐾_{23} =\frac{𝑉_{2}}{𝑉_{3}}=\frac{𝑉_{2}/{𝑉_{1}}}{𝑉_{3}/{𝑉_{1}}}=\frac{𝐾_{13}}{𝐾_{12}}}$$

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