Chapter 4 Additional Problems



Chapter 4 Additional Problems

X4.1 Determine the value of the coefficient of coupling (k) for the transformer of Example 4.7 of the text.

The turns ratio is

[pic]

Based on [4.27],

[pic]

By use of [4.25] and [4.26],

[pic]

[pic]

Using the conclusion of Problem 3.15,

[pic]

X4.2 For the ideal transformer circuit of Fig. X4.1, [pic], [pic], and [pic]. If [pic] and [pic], (a) determine the turns ratio a, (b) the source voltage [pic], and (c) the input power factor [pic].

(a)

[pic]

[pic]

[pic]

[pic]

(b)

[pic]

[pic]

[pic]

[pic]

(c)

[pic]

X4.3 For the circuit of Fig. X4.1, [pic], [pic], [pic], [pic], and [pic]. Calculate (a) [pic] and (b) [pic].

(a)

[pic]

[pic]

Assume [pic] on the reference.

[pic]

[pic]

[pic]

[pic]

(b)

[pic]

X4.4 For the circuit of Fig. X4.1, [pic], [pic], and [pic]. Determine the percentage of input power [pic] that is dissipated by [pic] regardless of the voltage values.

[pic]

[pic]

X4.5 For the circuit of Fig. X4.1, let [pic], [pic], [pic], and [pic]. Determine the power factor [pic].

With [pic], [pic] and [pic]; thus [pic] is shorted and the input impedance is [pic]. Since [pic] must lag [pic] by 90º,

[pic]

X4.6 The non-ideal transformer of Fig. X4.2 has three coils each with identical number of turns N. Terminal pair e-f is open circuit. If terminals b and c are connected together and [pic] (60 Hz), the input current is 1 A. If terminal pairs a-b and e-f are open circuit and [pic] (60 Hz), predict the value of input current. Leakage flux and coil resistance are to be neglected.

Based on [4.7], the maximum value of mutual flux for the two cases, respectively, is

[pic]

[pic]

Or, the mutual flux has the same value for both cases. Thus, the magnetizing current and core losses have identical values for both cases. Consequently, the input current for the second case must also be 1 A.

X4.7 The non-ideal transformer of Fig. X4.2 has three coils each with identical number of turns. Terminal pair e-f is open circuit. Leakage flux, coil resistance, and core losses can be neglected. When the two left-hand coils are additively connected in parallel and [pic] (60 Hz), the input current is 2 A. If the lower coil (terminal pair c-d) is disconnected (open circuit) and [pic] remains 120 V (60 Hz), predict the value of input current to the single coil.

Based on [4.7], the maximum value of mutual flux for both cases is given by

[pic]

Hence, the mmf established for both cases must be identical. For the first case of parallel-connected coils, the 2 A input current divides equally between the two coils to produce an exciting mmf of [pic]. When one of the two coils is disconnected, the current through the remaining coil must increase to 2 A to produce the required mmf of 2N. Thus, input current remains 2 A.

X4.8 For the ideal residential distribution transformer of Fig. X4.3, (a) determine current [pic]. (b) Assume that the two series-connected secondary windings are identical and determine the minimum kVA rating of a 2400:240/120 V transformer required to sustain this load without risk of winding over-temperature.

(a)

[pic]

[pic]

[pic]

(b) Since [pic] is the larger secondary current, the rating is dictated by the lower secondary winding; thus,

[pic]

X4.9 The transformer of Fig. X4.4 is rated as 3 kVA, 240:120 V, 60 Hz if [pic] is connected to [pic] with 240 V applied between [pic] and [pic]. However, it is connected as shown in Fig. X4.4 where [pic]. (a) Find the values of [pic] and [pic]. (b) Are all windings operating within rated current values?

(a) Conclude from the voltage rating that three coils have identical number of turns.

[pic]

[pic]

[pic]

(b) Letting R denote rated,

[pic]

[pic]

From the results of part (a), it is seen that the [pic] coil and the [pic] coil are operating within rated current.

[pic]

Thus, the current of the [pic] coil significantly exceeds the rated value of 12 A.

X4.10 Determine the value of [pic] for the ideal transformer of Fig. X4.5.

[pic]

[pic]

or

[pic]

X4.11 The three-winding ideal transformer of Fig. X4.6 has [pic] and identical load resistors (R) connected across coils 2 and 3. Determine the input impedance [pic] as indicated on Fig. X4.6.

MMF balance requires that

[pic]

[pic]

or

[pic]

Since the value of flux through all three coils is identical, [pic]. By Ohm's law,

[pic]

[pic]

Use (2) and (3) in (1) to find

[pic]

Hence,

[pic]

X4.12 A 15 kVA, 2400:240/120 V, 60 Hz, two-winding transformer is to be reconnected as a 2400:2520 V step-up autotransformer. From test work on the two-winding transformer, it is known that its rated voltage core losses and coil losses are 280 W and 300 W, respectively. For this autotransformer, (a) determine the apparent power rating and (b) the full-load efficiency if supplying 2520 V to a 0.8 PF lagging load.

(a) The connection is similar to Fig. 4.31b except that the upper coil consists of the parallel additive connection of the two 120 V secondary windings. Following the procedure of Example 4.14,

[pic]

[pic]

[pic]

(b) The core and copper losses are unchanged from the two-winding transformer.

[pic]

[pic]

X4.13 An autotransformer is frequently used as a variable voltage supply in the laboratory. The construction is a single coil wound on a toroidal core. A common lead exists between the input and output as shown by Fig. 4.32. The other output lead makes sliding contact with the coil. If the input voltage is impressed across the total span of the coil, then the two coil sections are always additive. If such an autotransformer is rated for 20 A output, what must be the current rating of the winding?

Let [pic] be the per unit portion of the N turn coil between output leads. Then mmf balance is given by

[pic]

Referring to Fig. 4.32,

[pic]

Simultaneous solution of (1) and (2) yields

[pic]

Whence it is seen that as [pic], [pic] and [pic]. Conversely, as [pic], [pic] and [pic]. Thus, the coil must be rated for 20 A to handle the extremes in output voltage.

X4.14 The transformer of Example 4.8 (using the approximate equivalent circuit) is supplying rated current and voltage to a load. For the load point, [pic]. Determine the load PF.

This described condition can only occur for a leading PF as illustrated by the phasor diagram of Fig. X4.7. By the Law of Cosines,

[pic]

From Example 4.8, [pic], and

[pic]

Then,

[pic]

By application of KVL,

[pic]

[pic]

X4.15 Three 100 kVA, 12,470:7200 V, 60 Hz, two-winding transformers are to be connected to provide 300 kVA, 7200/4160 V service to an industrial customer from the three-phase 7200 V distribution mains. (a) Determine the connection arrangement for this transformer bank. (b) For balanced operation, with an apparent power load of 150 kVA, determine the values of line current on both sides of the transformer bank.

(a) To meet the service agreement of 300 kVA with rated voltages, the transformers must be connected [pic] on the primary side. A wye connection on the secondary side provides for 7200 V line-to-line and 4157 V (nominal 4160 V) line-to-neutral.

(b) With the transformer bank operating at half rated apparent power capability and referring to Fig. 4.33d,

[pic]

[pic]

X4.16 A 25 kVA, 2400:240 V, 60 Hz, two-winding transformer is to be applied in a distribution system with service at 2000:200 V, 50 Hz. (a) Is there any fundamental problem with this application? (b) Determine the apparent power rating in the 50 Hz application.

(a) Based on [4.7], the maximum value of mutual flux is

[pic]

Since the ratio of voltage to frequency

[pic]

is unchanged, the magnetic core will operate at the design level of flux density with the magnetizing current unchanged. The core losses, being frequency dependent, will reduce in value, resulting in a cooler operating temperature. Operationally, there is no problem.

(b) The transformer can still only thermally handle the rated current for the 60 Hz design case. However, voltage has reduced. Thus,

[pic]

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