Holooly Plus Logo
Holooly Plus Logo

Question 3.26: A 440/220 V single-phase transformer has percentage resistan...

A 440/220 V single-phase transformer has percentage resistance drop and reactance drop of 1.2 per cent and 6 per cent, respectively. Calculate the voltage regulation of the transformer at 0.8 power factor lagging.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Percentage voltage regulation    =\frac{(I_{2}  R^{′′}_{e} \cos Φ  +  I_{2}  X^{′′}_{e} \sin Φ)  }{E_{2}} ×  100

 

= \frac{ I_{2}  R^{′′}_{e} }{ E_{2}} \cos Φ  ×   100 + \frac{ I_{2}  X^{′′}_{e} }{ E_{2}} \sin Φ  ×   100.

Given,                                        \frac{ I_{2}  R^{′′}_{e} }{ E_{2}}  ×   100 = 1.2, \frac{ I_{2}  X^{′′}_{e} }{ E_{2}}  ×   100 = 6.0,  \cos Φ = 0.8

 

\sin Φ = 0.6.

Substituting values regulation                  = \frac{ I_{2}  R^{′′}_{e} }{ E_{2}}  \cos Φ  + \frac{ I_{2}  X^{′′}_{e} }{ E_{2}} \sin Φ

= 1.2 ×.0.8 + 6.0 × 0.6
= 4.56 per cent

Related Answered Questions

Question: 3.24

The equivalent circuit parameters of a 300 kVA, 2200/200 V, 50 Hz single-phase transformer are: primary winding resistance, R1 = 0.1 Ω; secondary winding resistance, R2 = 0.01 Ω; primary leakage reactance, X1 = 0.4 Ω; secondary leakage reactance, X2 = 0.03 Ω; resistance representing core loss, ...

Verified Answer:

We will consider E_{1} = 2200  V an...
Question: 3.25

A 10 kVA 440/220 V, 50 Hz single-phase transformer gave the following test results when both the following tests were conducted on the high-voltage side: Open circuit test : 440 V, 1.0 A, 100 W Short circuit test : 20 V, 22.7 A, 130 W. Using the test data, calculate the efficiency and voltage ...

Verified Answer:

Full-load current on the high-voltage side = [late...
Question: 3.27

Calculate the all-day efficiency of a 25 kVA distribution transformer whose loading pattern is as follows: 15 kW at 0.8 power factor for 6 hours 12 kW at 0.7 power factor for 6 hours 10 kW at 0.9 power factor for 8 hours Negligible load for 4 hours. The core loss is 500 W and full-load copper loss ...

Verified Answer:

Output of the transformer in 24 hours is calculate...
Question: 3.6

A parallel magnetic circuit with 2000 turns on its central limb has been shown in Fig. 3.18. The air gaps are 2 mm each. The mean diameter of the circular magnetic path is 20 cm. The cross-sectional area of the central limb is 10 cm² while the cross section of the outer limbs is 5 cm². A few ...

Verified Answer:

The flux from the central limb will get equally di...
Question: 3.23

Efficiency of 400/200 V, 200 kVA transformer is 98.5 per cent at full load at 0.8 lagging power factor. At half load, 0.8 power factor lagging the efficiency is 97.5 per cent. Calculate the values of core loss and full-load copper loss. ...

Verified Answer:

Let W_{c} be the core loss and [la...
Question: 3.14

A transformer has 900 turns on its primary winding and 300 turns on its secondary. A voltage of 230 V at 50 Hz is connected across its primary winding. The cross-sectional area of the core is 64 cm². Calculate the magnitude of the induced EMF in the secondary winding. Also calculate the value of ...

Verified Answer:

Given V_{1} = 230  V. The induced ...
Question: 3.18

The no-load input power to a transformer is 100 W. The no-load current is 3 A when the primary applied voltage is 230 V at 50 Hz. The resistance of the primary winding in 0.5 Ω. Calculate the value of iron loss and no-load power factor. ...

Verified Answer:

At no-load, the input power is lost as a small amo...
Question: 3.17

The maximum flux density in the core of a 1100/220 V, 50 Hz, 100 kVA transformer is 3.5 Wb/m². Calculate the area of cross section of the core and the number of turns of the primary and secondary windings if the EMF per turn is 5.5 V. ...

Verified Answer:

V_{1} = E_{1} = 1100  V EMF per tur...
Question: 3.12

There is mutual magnetic coupling between two coils of number of turns 500 and 2000, respectively. Only 50% of the flux produced by the coil of 500 turns is linked with the coil of 1000 turns. Calculate the mutual inductance of the two coils. Also calculate the EMF induced in the coil of 1000 turns ...

Verified Answer:

Mutual inductance, M=\frac{flux  linkage  ...
Question: 3.11

An iron-cored toroidal coil has 100 turns. The mean length of the flux path is 0.5 m an the cross-sectional area of the core is 10 cm². Calculate the inductance of the coil. Assume relative permeability of iron as 2000. Also calculate the induced emf in the coil when current of 5 A is reversed in ...

Verified Answer:

The expression for inductance in terms of its para...