EXCEL FIRST REVIEW AND TRAINING CENTER, INC.
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Answers & Solutions:
1. C. The effective value is the same
as that of the original ac wave.
Elec 09-01 to Elec 09-03
Power Supply
14. B. A quick-break type
32. C. The filter
15. C. Pulsating dc
33. A. 16 V, –4 V
16. A. 25:1
2. A. A capacitor in parallel with the
rectifier output.
3. D. A capacitor in parallel and a
choke in series
4. C. The output is hard to filter
5. C. The filter capacitor(s) must be
initially charged
Solution:
2
 N1 
R L ' 10 k

 625

 
RL
16 
 N2 
N1
 625  25 :1  Ans.
N2
Solution:
During the positive half-cycle:
Diode-Reverse Bias
6. C. A bleeder resistor
17. D. A capacitor in parallel with the
dc output
7. B. 15.7 Vrms
18. B. A TV broadcast transmitter
Solution:
VP N P

Vs Ns
220 Vrms 14

Vs
1
Vs  15.7 Vrms  Ans.
19. D. Use two capacitor/choke
sections one after the other
20. A. It uses the whole transformer
secondary for the entire ac
input cycle
21. A. The primary voltage is greater
than the secondary voltage
22. C. Has one tapped winding
9. C. 180 degrees
23. B. Half of the ac wave is chopped
off.
10. A. twice that from a half-wave
circuit
24. A. Provides maximum coupling
13. B. 1.45 A, 10.09 %
28. C. low-pass filter
Solution:
29. B. Full wave, center-tap
V 10.9 V

R 7.5 
I L  1.45 A  Ans.
30. B. The I2R losses are lower
IL 
VR 
VNL  VFL
x100%
VFL
12  10.9
x100%
10.9
VR  10.09 %  Ans.
VR 
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34. B. There will be an impedance
mismatch, no matter what the
turns ratio of the transformer
36. C. The fuse will blow out right
away
37. B. A step-down unit
40. D. Poor regulation
41. B. The effective value is less than
that of the original ac wave.
42. D. 53.6 V
Solution:
31. C. 40:1
Vdc 
Solution:
NP

NS
Vo   4 V
39. C. The primary voltage is less than
the secondary voltage
26. C. capacitor
27. A. Diode failure
Vo  0  4V
38. C. Low-pass filter
25. D. The balanced winding
12. A. pulsating dc voltage.
During the negative half-cycle:
Diode-Forward Bias
35. B. rectifier
8. D. All of the above are generally
needed
11. B. Parallel with the filter output,
reverse-biased
I0
Vo  16 V
ZP
12.8 k 40


 Ans.
ZS
8
1
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RL
Vdc
R  RL
1000 
 60 V 
120   1000 
V 'dc  53.6 V  Ans.
Vdc 
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EXCEL FIRST REVIEW AND TRAINING CENTER, INC.
Cebu: JRT Bldg. Imus Ave, Cebu City | Davao: 2nd Fl. MERCO Bldg. Rizal St. cor. Bolton St.
Manila: CMFFI Bldg. R. Papa St. Sampaloc | Baguio: 4th Fl. De Guzman Bldg., Legarda Road
Elec 09-01 to Elec 09-03
Power Supply
43. C. An increase in core loss
44. A. Connected in parallel with filter
capacitors
45. C. 2
46. A. Placing the windings on
opposite sides of a toroidal core.
47. B. 16.96 V
Solution:
Vdc  Vp 
Idc
4fC
 18 V 
100mA
4  60Hz  400F 
 16.96 V  Ans.
48. C. ripple
49. A. 63.6 V
Solution:
Vdc  0.318Vm
Vdc  0.318  200 
Vdc  63.6 V  Ans.
50. C. Leave it alone and have a
professional work on it
END
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