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352 Test: Part two 30. The vector 40 + j30 represents: A. 40 resistance and 30 H inductance. B. 40 uH inductance and 30 resistance. C. 40 resistance and 30 inductive reactance. D. 40 inductive reactance and 30 resistance. E. 40 uH inductive reactance and 30 resistance. 31. In a series RC circuit, where, R 300 and XC 30 : A. The current leads the voltage by a few degrees. B. The current leads the voltage by almost 90 degrees. C. The voltage leads the current by a few degrees. D. The voltage leads the current by almost 90 degrees. E. The voltage leads the current by 90 degrees. 32. In a step-down transformer: A. The primary voltage is greater than the secondary voltage. B. The primary impedance is less than the secondary impedance. C. The secondary voltage is greater than the primary voltage. D. The output frequency is higher than the input frequency. E. The output frequency is lower than the input frequency. 33. A capacitor of 470 pF is in parallel with an inductor of 4.44 H. What is the resonant frequency A. 3.49 MHz. B. 3.49 kHz. C. 13.0 MHz. D. 13.0 GHz. E. Not determinable from the data given. 34. A sine wave contains energy at: A. Just one frequency. B. A frequency and its even harmonics. C. A frequency and its odd harmonics. D. A frequency and all its harmonics. E. A frequency and its second harmonic only. 35. Inductive susceptance is: A. The reciprocal of inductance. B. Negative imaginary. C. Equal to capacitive reactance. D. The reciprocal of capacitive susceptance. E. A measure of the opposition a coil offers to ac.

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In Chap. 19, you learned that a diode will not conduct until the forward-bias voltage is at least as great as the forward breakover voltage. There s a corollary to this: a diode will always conduct when the forward-bias voltage reaches or exceeds the forward breakover voltage, when the device is conducting current in the forward direction. In the case of silicon diodes this is approximately 0.6 V. For germanium diodes it is about 0.3 V, and for selenium diodes it is about 1 V. This phenomenon can be used to advantage when it is necessary to limit the amplitude of a signal, as shown in Fig. 20-8. By connecting two identical diodes back-to-back in parallel with the signal path (A), the maximum peak amplitude is limited, or clipped, to the forward breakover voltage of the diodes. The input and output waveforms of a clipped signal are illustrated at B. This scheme is sometimes used in radio receivers to prevent blasting when a strong signal comes in. The downside of the diode limiter circuit, such as the one shown in Fig. 20-8, is the fact that it introduces distortion when clipping occurs. This might not be a problem for reception of digital signals, for frequency-modulated signals, or for analog signals that rarely reach the limiting voltage. But for amplitude-modulated signals with peaks that rise past the limiting voltage, it can cause trouble.

Test: part two 353 36 The rate of change (derivative) of a sine wave is itself a wave that: A Is in phase with the original wave B Is 180 degrees out of phase with the original wave C Leads the original wave by 45 degrees of phase D Lags the original wave by 90 degrees of phase E Leads the original wave by 90 degrees of phase 37 True power is equal to: A VA power plus imaginary power B Imaginary power minus VA power C Vector difference of VA and reactive power D VA power; the two are the same thing E 0707 times the VA power 38 Three capacitors are connected in series Their values are 47 F, 68 F, and 100 F The total capacitance is: A 215 F B Between 68 F and 100 F C Between 47 F and 68 F D.

chromatic dispersion in the fiber; the more chromatic dispersion, the more SPM. It is counteracted with the use of large effective area fibers.

two diodes to act as an ac limiter. At B, illustration of sinewave peaks cut off by the action of the diodes in an ac limiter.

22 F E Not determinable from the data given 39 The reactance of a section of transmission line depends on all of the following except: A The velocity factor of the line B The length of the section C The current in the line D The frequency E The wavelength 40 When confronted with a parallel RLC circuit and you need to find the complex impedance: A Just add the resistance and reactance to get R + jX B Find the net conductance and susceptance, then convert to resistance and reactance, and add these to get R + jX C Find the net conductance and susceptance, and just add these together to get R + jX D Rearrange the components so they re in series, and find the complex impedance of that circuit E Subtract reactance from resistance to get R jX 41 The illustration in Fig.

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