B- what are the disadvantages of the auto lran
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![B- what are the disadvantages of the auto transformers?
C- An auto transformer is used to step-down from 240V to 200V. the complete winding consists
of 438 turns and the secondary delivers a load current of 15A. determine (1) secondary turns (2)
primary current (3) current in the secondary winding and (4) the saving of copper.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff9f0c0fd-03d6-490a-a851-ebdb721488a1%2F1b7dfe97-6e72-4fa3-b902-b29d5dc9796c%2Flni08lm_processed.jpeg&w=3840&q=75)
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- Two transformers A and B have the following data Transformer A: 6600/440 V, (250 KVA ) requires ( 6%) of rated voltage to circulate (90%) of full load current at power factor (0.25) when the low voltage winding is short circuited and the no load loss is equal to (2500 W). Transformer B: 6600/420 V, (600 KVA) requires (9%) of rated voltage to circulate (80%) of full load current at power factor (0.16) when low voltage winding is short circuited and no load loss is equal to (4500 W). Calculate the followings: If both transformers are connected in parallel to supply a load (1.2 + j 2.3) ohm, find the output power of each transformer.Two transformers A and B have the following data Transformer A: 6600/440 V, (250 KVA ) requires ( 6%) of rated voltage to circulate (90%) of full load current at power factor (0.25) when the low voltage winding is short circuited and the no load loss is equal to (2500 W). Transformer B: 6600/420 V, (600 KVA) requires (9%) of rated voltage to circulate (80%) of full load current at power factor (0.16) when low voltage winding is short circuited and no load loss is equal to (4500 W). Calculate the followings: The terminal voltage when transformer A is supplying half load at power factor (0.6) lagging and transformer B is supplying (85%) of full load at power factor (0.85) leading.1.a. For performing dielectric-strength tests on insulating materials, two transformers are connected in aA. parallel combination.B. cascade arrangement.C. polyphase arrangement. 1.b. In large power transformers, the unequal voltage distribution caused by voltage surges in the winding isprevented by the use ofA. static shields.B. series capacitors.c. low-voltage arresters.
- In the circuit in Figure 2, the capacitor connected to the a-b terminals of the transformer anda circuit consisting of resistance is defined as a load circuit. a)Find the average power spent by the load circuit? b) How much of the average power calculated in a) by the resistance in the load circuit and how muchspent by the capacitor? Comment on the results obtained?3. Why the copper loss in the primary can be neglected, in the open circuit test? 4. Why the terms Rm and Xm significant when the transformer is loaded?In the single-phase, three- winding, transformer, the primary current is equal to lo at the condition of ints there is no load on tertiary windings the current in secondary windings equal to zero the resistances (R1, R2, and R3) in transformer windings are equal the inductances (L1, L2, and L3) in transformer windings are equal.
- Q Two transformers A and B have the following data Transformer A: 6600/440 V, (250 KVA ) requires ( 6%) of rated voltage to circulate (90%) of full load current at power factor (0.25) when the low voltage winding is short circuited and the no load loss is equal to (2500 W). Transformer B: 6600/420 V, (600 KVA) requires (9%) of rated voltage to circulate (80%) of full load current at power factor (0.16) when low voltage winding is short circuited and no load loss is equal to (4500 W). Calculate the followings: The maximum efficiency of each transformer at power factor (0.75) lagging.The ratings of a single-phase three-winding transformer are as follows:*primary winding: 300 MVA, 14kV*secondary winding: 300MVA, 200kV*tertiary winding: 40MVA, 20kV. The leakage reactances are:*Xps = 0.1 at 300MVA, 14kV*Xpt = 0.16 at 40MVA, 14kV*Xst = 0.14 at 50MVA, 200kV Neglecting the winding resistance and exciting current, calculate Xp, Xs and Xt using 300 MVA and 14kV as the base for the primary winding.A voltage transformer in Figure 8 has 1,500 turns of wire on its primary winding and 500 turns of wire for its secondary winding. What will be the turns ratio, a of the transformer. If 240 volts is applied to the primary winding, what will be the resulting secondary voltage. And if a 400 n resistor is connected to the secondary winding, what are the values of current Is and Ip. primary Vp Transformer Core NF Ns Figure 8 secondary Vs
- A single-phase transformer is required to step down the voltage from 1100 V to 400 V at 50 Hz. The [5]core has a cross-sectional area of 25 cm? and the maximum flux density is 5Wb/m?. Determine thenumber of turns of the primary and secondary windings.Flag this question for future reference What is the test that can be conducted in the laboratory to understand the transformer impedance drops are dependent on the load current with a variation in the secondary voltage? O a. Direct Load Test O b. Open Circuit Test OC. Voltage Ratio Test O d. Polarity TestB- what are the disadvantages of the auto transformers? C-An auto transformer is used to step-down from 240V to 200V. the complete winding consists of 438 turns and the secondary delivers a load current of 15A. determine (1) secondary turns (2) primary current (3) current in the secondary winding and (4) the saving of copper.
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