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Cram for The PE Sample Test IV

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1 hour ago, Cram For The PE said:

A full wave rectified circuit is given below.
Untitled.png.4a06d8afc705d4ed8f4d4cf2d55474c0.png

If diode D2 is shorted out, what is the average voltage across RL?

A) 0 (i.e. short circuit)
B) 54 V
C) 108 V
D) 216 V

 

http://cramforthepe.com/index.php/2020/02/08/power-electronic-problem-2/

 

I would say "A".  No current would flow through resistor RL. Here is re-drawing the circuit with the short and single equivalent circuit.

main-qimg-4b9514650196a8cf22c60da104bea702.jpeg

main-qimg-f08200b8311a443058db23f357b4e1e9.jpeg

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I think B.

The circuit above essentially acts like a half-wave rectifier because of the shorted diode. The average voltage would be Vavg=0.318*Vpk. With the two 10ohm resistors, the voltage is divided by 2. So peak voltage across load, Vpk = 240*√2 / 2 = 169.7. Vavg = 169.7*0.318=~54V.

Edited by Chattaneer PE

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B

during + swing

2 x sqrt2/Pi x 240/20 x 10 = 108

during - swing

RL shunted so RL sees no i so V = 0

avg = 54

Edited by Dude99

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ME to EE, your two graphs that you drew do no match, which is likely why you arrived at the wrong answer.

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57 minutes ago, a4u2fear said:

ME to EE, your two graphs that you drew do no match, which is likely why you arrived at the wrong answer.

Thank you.  I also did some research and found some more half-wave and full-wave rectifier problems to work on so I could better understand them.  I realized I had not spent enough time on this types of problems where you have to consider both cases (positive and negative) of current direction.  

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On 2/8/2020 at 5:37 PM, Chattaneer PE said:

I think B.

The circuit above essentially acts like a half-wave rectifier because of the shorted diode. The average voltage would be Vavg=0.318*Vpk. With the two 10ohm resistors, the voltage is divided by 2. So peak voltage across load, Vpk = 240*√2 / 2 = 169.7. Vavg = 169.7*0.318=~54V.

Will the answer be the same if diode D2 will be open and not shorted?

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1 hour ago, BebeshKing said:

Will the answer be the same if diode D2 will be open and not shorted?

Yes, it basically acts like a half-wave rectifier again. However, in the case of D2 being open, the negative part of the sign wave is rectified. Whereas in the case of D2 being shorted, the positive portion is passed through.

Excuse my crude diagrams:

Green arrows are the positive portion of the cycle, and red is the negative portion of the cycle. Notice with the regular circuit (full-wave bridge rectifier), current during both positive and negative portions of the cycle flows through the resistor RL. For the second pair of diagrams, current only flows through the resistor during the positive portion, therefore there is only voltage across the resistor during the positive portion. In the third set of diagrams, current only flows through the resistor during the negative portion of the cycle, therefore there is only voltage across the resistor during the negative portion. The third case is different since no current flows when the source voltage is positive.

image.png.f3b77827218869dbe01625ee4035fd73.png

 

And for future reference since we know we're now dealing with a half-wave rectifier in either case, the average (mean) is:

image.png.75d0e2360f7fdea59717a2366607ffca.png

Edited by Chattaneer PE
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