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CHAPTER 10 Magnetic Coupling. Transformers
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" where Zref impedance coupled or re ected into the primary circuit from the secondary " " circuit (note that Zref appears in series with the primary coil); and Z2 total series impedance of the secondary circuit, considered by itself, as de ned following eq. (383). Hence, as far as the generator is concerned, Fig. 222 can be redrawn as in Fig. 223.
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Thus, from Fig. 223, " V " I1 " " Z1 Zref Another useful relationship is found by solving eq. (385) for I2 ; thus " j!M I1 " I2 "2 Z 388
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Problem 186 In Fig. 222 both current arrows are drawn in the clockwise sense. What change, if any, would appear in eq. (389) if (a) (b) both arrows were drawn in the counterclockwise sense " " if the I1 arrow remained in the cw sense but the I2 arrow were drawn in the ccw sense
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Problem 187 In Fig. 224, the primary and secondary coils have equal inductances of 45 millihenrys. (Any resistance the coils may have is included in the 5-ohm and 2-ohm resistance values.) Find the magnitude of generator current, given that the frequency is 200 rad/sec. (Answer: 5.249 amperes)
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(See discussion note given with the solution to the above problem.) Problem 188 In Fig. 225, a generator of zero internal impedance produces 28 volts at 400,000 rad/ sec. The primary and secondary coils have equal inductances of 60 microhenrys.
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CHAPTER 10 Magnetic Coupling. Transformers
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Fig. 225
Find, (a) (b) magnitude of generator current, magnitude of current in secondary coil. (Answer: 6.030 amperes) (Answer: 6.000 amperes)
Problem 189 In Fig. 226, the generator frequency is 100,000/2 hertz (Hz), the values of the circuit parameters being in ohms, microhenrys, and microfarads. It is also given that L1 120 mH and L2 200 mH.
Fig. 226
(a) (b)
Find the magnitude of generator current. Find the magnitude of secondary current.
(Answer: 10.966 amperes) (Answer: 7.292 amperes)
Parts (c) through (e), which follow, will serve as a review of some important points concerning the calculation of POWER, as developed in section 8.5. (c) Using only the details found in part (a) and the principle developed in connection with Fig. 150, nd the TRUE POWER produced by the generator in Fig. 226. Repeat part (c), now making use only of eq. (228) and Fig. 155. Repeat part (c), now making use of the fact that the power produced in a pure resistance is equal to the square of the magnitude of current, times resistance.
(d) (e)
In problems 187, 188, and 189 it should be noted that we did not need to dot-mark the transformers in order to nd the required answers. This is because our analysis was based upon Fig. 222, which led to the basic pair of eqs. (384) and (385). Let us consider this in more detail, as follows. Suppose, in Fig. 222, that the position of the dots on the secondary side had been reversed (everything else remaining the same). In that case eqs. (384) and (385) would
have read
CHAPTER 10 Magnetic Coupling. Transformers
" " " " Z1 I1 j!M I2 V " " j!M I1 Z2 I2 0 " which, you ll notice, will give the same value of I1 as given by eq. (386) (and the same value of re ected impedance as given by eq. (387)). As a matter of fact, changing the position of the dots in Fig. 222 would only have the e ect of changing the phase angle of the second" ary current by 180 degrees (the same as changing the direction of I2 , as discussed in problem 186). It must be emphasized, however, that in some types of circuit it is necessary to take the placement of the dots into account. These are cases in which the primary and secondary currents are not totally separated, as in Fig. 222, but, instead, share a common circuit element. In such cases we must be careful to assign the proper algebraic sign to all mutually " induced voltages of the form j!M I . This is illustrated in the following problems. Problem 190 Figures 227 and 228 depict two circuits that are identical in all respects EXCEPT for " the placement of the reference dots. Write the equation for the current I for (a) Fig. 227, (b) Fig. 228.
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