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CHAPTER 10 Magnetic Coupling. Transformers
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Now suppose, in Fig. 244, that a sinusoidal voltage of V reference volts is applied to the " input terminals with the secondary side open-circuited. Then Zin jX1 , and the input current that ows in this condition is called the transformer magnetizing current, " which we ll denote as Im . By Ohm s law, " Im V=jX1 j V=X1 showing that, since X1 is very large, the magnetizing current of a high-quality iron-core transformer is a VERY SMALL CURRENT, LAGGING THE APPLIED VOLTAGE V BY 90 DEGREES. Recall that true power P in an ac circuit is, by eq. (227) of Chap. 8, equal to P VI cos , where  is the phase angle between V and I. Since for the above condition  908, and since cos 908 0, we see that the magnetizing current consumes NO ENERGY, and is thus spoken of as the wattless magnetizing current. In this regard, let us note that in a high-quality iron-core transformer the amount of magnetic ux in the iron core, and the amount of magnetizing current, both remain very nearly constant in value, independent of the amount of load current drawn by ZL . This is because when alternating current ows in the secondary winding it tends to oppose or buck the alternating ux produced in the core by the primary current (this is in accordance with Lenz s law, section 7.4). This reduction of alternating ux causes a lower counter emf to be induced into the primary coil, which at once allows more current to ow into the coil, bringing the ux back up to its previous level. To continue on, let us note that the important eq. (417) is expressed in terms of the constant a, de ned as a X1 !L1 L1 X2 !L2 L2
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which shows that eq. (417) is basically expressed in terms of the ratio of the inductances of the primary and secondary coils. Actually, however, in practical work (dealing with iron-core transformers only) it s much more convenient to deal with TURNS RATIO than inductance ratio. The turns ratio T of a transformer is de ned as T number of turns of wire on primary coil N 1 number of turns of wire on secondary coil N2
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It thus follows that, in order to express eq. (417) in terms of turns ratio we must know the relationship that exists between the inductance L of a coil and the number of turns N the coil has. Fortunately, for the ideal case of k 100% it is known that the INDUCTANCE of an ideal coil is proportional to the SQUARE of the number of turns, that is, L k 0N 2 where k 0 is a constant of proportionality.* Thus, since the two similarly constructed coils of Fig. 244 would have the same value of k 0 , it would be true that
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2 L1 k 0 N1
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2 L2 k 0 N2
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and hence, upon substituting these values in the above equation for a, we have that
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2 2 a N1 =N2 N1 =N2 2 T 2
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* See note 26 in Appendix.
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CHAPTER 10 Magnetic Coupling. Transformers
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and thus eq. (417) can be written in the more practical form " " Zin T 2 R jX T 2 ZL 418
where T is the ratio of PRIMARY TURNS TO SECONDARY TURNS. Another important fact can be deduced as follows. Since k 100% in the ideal case, this means that the same value of ux, created by the magnetizing current, links every turn of the transformer. Hence the volts induced per turn is the same on both the primary and secondary sides. Thus, if V1 is the voltage across the N1 primary turns and V2 is the voltage across the N2 secondary turns, then V1 V2 N1 N2 that is, V1 N1 T V2 N2 419
showing that, in an ideal transformer, the VOLTAGE RATIO is equal to the TURNS RATIO. It is also true that the power input and the power output of any type of transformer are equal to Pin V1 I1 cos 1 and Pout V2 I2 cos 2 . In an ideal transformer, however, 1 2 and hence, in an ideal transformer, V1 I1 V2 I2 that is, V1 I2 V2 I1 420
Problem 199 A load impedance of (3 j5) ohms is connected to the secondary terminals of an ideal transformer of turns ratio 4 to 1, primary to secondary. A constant value of 240 volts rms is applied to the primary terminals. Find the following: (a) impedance seen looking into primary terminals, (b) secondary current, (c) power to load. Problem 200 A transformer has a primary inductance of L1 henrys and a secondary inductance of L2 henrys, with coe cient of coupling k. Assuming negligible winding resistance, show that, if the secondary terminals are shorted together, the impedance seen looking into the primary terminals would be equal to " Zin j!L1 1 k2 ohms Problem 201 An iron-core power transformer has primary inductance of 4 henrys and secondary inductance of 3 henrys, with 99.5% coe cient of coupling. The windings have negligible resistance. If the transformer is connected to a 120 volt, 60 Hz power line, how much line current would theoretically ow if the secondary terminals were accidentally shorted together
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