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Since we desire that no energy be lost in the T network itself, it follows that the three elements of the network will have to be composed of pure reactances only. As a rst possibility, suppose we made all three elements of the T network inductive reactances (coils), as in the gure below.
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In such a case it s apparent that we could never see a pure resistance looking into (1, 1) because there is no capacitance present to cancel out the inductive reactance. In the same way, if all three elements were capacitors we could not possibly see a pure resistance looking into terminals (1, 1). Thus, for the particular conditions of this section, the T network will have to contain both inductive reactance and capacitive reactance only then can there be a total cancellation of reactances, making it possible to see a pure resistance, Rin , when looking into terminals (1, 1). Thus we must have either two inductors and one capacitor, or two capacitors and one inductor, as shown in Figs. 183 and 184.
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Fig. 183
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In Fig. 183 note that Z1 jX1 , Z2 jX2 , Z3 jX3 , while, in Fig. 184, Z1 jX1 , Z2 jX2 , Z3 jX3 . Putting these values into eq. (108) we have, for Fig. 183: RRin jRin X2 X3 X1 X2 X1 X3 X2 X3 jR X1 X3 for Fig. 184: RRin jRin X2 X3 X1 X2 X1 X3 X2 X3 jR X1 X3 Recall that two complex numbers can be equal only if the two real parts are equal and the two imaginary parts are equal; hence, inspection of the last two equations shows that for both Figs. 183 and 184: RRin X1 X2 X1 X3 X2 X3 for Fig. 183: Rin X2 X3 R X1 X3 310 309
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for Fig. 184:
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Rin X2 X3 R X1 X3 311
In a practical problem the value of the actual load resistance R (Fig. 182) and the desired value of Rin would be known. This would leave us with three unknown reactances and, for either of the above two gures, just two simultaneous equations. In such a case we could select any reasonable value for one of the three reactances, say X3 , and then use the two equations to calculate the required values of the other two reactances. Actually, however, in most practical work a balanced T (or balanced pi if we re using a pi network) network would be used. A balanced network is one in which the magnitudes of the three reactances are all equal in value; thus, setting X1 X2 X3 X in eq. (309), we nd that, for a balanced T network, the common magnitude of reactance, for either Fig. 183 or Fig. 184, is equal to p 312 X RRin In addition to the balanced T network, the balanced pi network, shown in Fig. 185, nds wide use as the output stage of a radio transmitter.
Fig. 185
In terms of standard pi-network notation (Fig. 175 in section 9.2) we have, for the balanced pi network of Fig. 185, " " " ZA jX; ZB jX; ZC jX We can, if we wish, convert the above pi network into an equivalent T network, thus by eq: 302 ; by eq: 303 ; by eq: 301 ; jX jX " jX Z1 jX jX jX " jX Z2 jX jX jX " jX Z3 jX
We can now make use of the equations for the T network, and thus, setting X1 X2 X3 X in eq. (309), we have, for the balanced pi network of Fig. 185, that p 313 X RRin as for the balanced T network (eq. (312)). Problem 162 A generator, having an internal resistance of 36 ohms, generates, on open circuit, 90 volts rms at a frequency of 175 kHz (kilohertz). It is necessary that the generator
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