The question asks, what is the average value of the output voltage fro the circuit. The choices are; 5.48v 2.74v 0v and 6.37v
I choose 6.37v using a simple formula; 2(160)/pie then divid by 16. However, the book says the answer is 5.48v. Could the book have made an error with their answer?

 

You have to account for about 0.6v drop across the rectifier diode.

 

You have to account for about 0.6v drop across the rectifier diode.

I also have tried that. Our text uses .7 for Silicon but anyway, it still doesn't equal. 5.48v

 

This is a trick question.
The question asks for average value, not the rms value.

 

I don't think its asking for rms or avg. If I use avg which is .637 the answer would come to 4.05v. Not the first time the book has mad an error

 

questions:

1. What is the output of the transformer?
2. What is the peak signal on the load?
3. What is the average respresentated by that peak?

The book is correct ....

 

Remember that each half cycle that appears across the load has already had the drop from two diodes subtracted from it,so that is all you get to work with.

 

See the attached image.

The voltage across the load resistor doesn't look like the blue curve; it looks like the red curve (both waveforms have peak values of 8.6 volts).

The average of the blue voltage waveform is 5.47493 and the average of the red waveform is 5.02869

I think none of the book answers are correct.

 

Out of the transformer comes a 10v peak waveform fully rectified half cycles, at any time during conduction the diode drop is approx 0.7v*2 = 1.4v so it's almost a 8.6v peak waveform (apart from some reduction in the total ON period) so the average is less than 8.6 * 0.707 = 6.08v.

I don't know offhand how much the conduction angle affects the voltage but it is definitely less than 6.08v, so the best answer would be the next value down which is 5.48v. (Because 2.74 is definitely too low, and 6.37 is definitely too high).

 

Out of the transformer comes a 10v peak waveform fully rectified half cycles, at any time during conduction the diode drop is approx 0.7v*2 = 1.4v so it's almost a 8.6v peak waveform (apart from some reduction in the total ON period) so the average is less than 8.6 * 0.707 = 6.08v.*

I don't know offhand how much the conduction angle affects the voltage but it is definitely less than 6.08v, so the best answer would be the next value down which is 5.48v. (Because 2.74 is definitely too low, and 6.37 is definitely too high).

* That's not average,it's RMS.

Average is 8.6* 0.637=5.478≈5.48v

 

See the attached image.

The voltage across the load resistor doesn't look like the blue curve; it looks like the red curve (both waveforms have peak values of 8.6 volts).

The average of the blue voltage waveform is 5.47493 and the average of the red waveform is 5.02869

I think none of the book answers are correct.

Yeah,you have allowed for the fact that the "half cycles" now have time gaps between them,so the average falls.
Come to think of it,they now have a shorter rise time compared with their repetition rate,so that would change the value of the average figure too.
It seems like the book just took the average of one half cycle,assuming sinewave shape.

My head hurts!

 

The Electrician is being pedantic but correct.

 

* That's not average,it's RMS.

Average is 8.6* 0.637=5.478≈5.48v

But the voltage out of a full wave rectifier into a resistive load is not an average it's an RMS. I assumed "average" was the OP's wording choice.

 

When the OP identified the correct answer as 5.48 ... I knew they did the three steps I outlined.

Of course, the Electrician set us all straight on how to really solve it for the average voltage.

 

But the voltage out of a full wave rectifier into a resistive load is not an average it's an RMS.

What does it mean to say that a voltage "is" or "is not" average or RMS? How can a voltage have the property of being RMS? Isn't it just a matter of how one chooses to measure it? Can't I measure a voltage with a meter that indicates the average (DC) value, or the RMS value, as I choose, or as some requirement compels me to do?

If one cares about the heating of the resistive load, then the RMS value is appropriate. If, on the other hand, the resistor is high valued and not much of the power is being dissipated in it, but most of the current is going to charge a battery for example, then the average value is appropriate.

I assumed "average" was the OP's wording choice.

The very first sentence in post #1 says "The question asks, what is the average value of the output voltage fro the circuit." That suggests to me that the problem is asking for average value rather than it being the OP's choice of wording.

 

Well generally in full wave PSU design we care about the power output, so RMS calcs are the norm. I don't ever remember seeing a full wave PSU question in a test asking about "averages", but then it's been 30 years since I sat in the classroom so I should probably shut up now!

 

Well generally in full wave PSU design we care about the power output, so RMS calcs are the norm. I don't ever remember seeing a full wave PSU question in a test asking about "averages", but then it's been 30 years since I sat in the classroom so I should probably shut up now!

Expect the unexpected in the Homework Section ... but keep participating.