ok, i have a cocky nephew who thinks his college courses are the hardest out there. im not denying electronics is hard, as a former law student i know how tough college can be. But as a young kid he needs to learn how to word his bets carefully. In law school wording is everything, so as the older wiser uncle i need to teach him a lesson. He bet his new mustang for a month for the keys to my house for a month that i could not find the answer to this electronics question. I dont have to show my work or even tell him how i got the answer, only stipulation is i cant ask a professor. So if your a professor dont answer lol. here it goes

i have to find the average current running through a circuit, with a DC power source in series with a resister... an ideal diode... and an AC power source.
The AC source is 16cos(400t) DC is 13 volts and the resister is 2 ohms

sorry if this is not properly worded but thats what he wrote down for me. If someone can help me out id really appreciate it, and for all you guys out there doing these kind of problems the headache medicine is on me. Thank god people out there actually like this stuff because i wouldnt have all my gadgets without ya lol. later

 

hyedenny

How "wise" is it to set an example for a "young kid" by cheating?
Did your tough law school curriculum include classes on dishonesty and deception? Do you frequently cheat your family? Does the youngster make you feel inferior? desperate? dumb?

Typical lawyer... I hope the brat trashes your house when he wins! Meanwhile, Im going to figure out the answer to your question on my own, although I think you left out either a frequency or time component.

I think you proved his point.

 

Frequency of the AC source is 400 radians per second. Divide by 2*pi to get the frequency in Hz.

I like 6.54 Amps for a first cut.

 

But why would this be cheating? Do you really think the nephew expected him to learn electronics from scratch? Actually, it's just a trig and calculus problem if you know what an ideal diode is.
I get 6.70A or 196.9mA, depending on which way the diode is installed.
BTW, I also "cheated". I simulated it. It's much easier to have Spice do your trig and integration for you. 8)

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Ron

 

I don't think he's cheating either, but now he has three different answers to choose from. Who ya gonna believe.

BTW in the simulation was there any conduction on the negative half cycle?

 

This is a tough one. An AC power source provides an output at a fixed frequency. Our sockets in our walls (americans and canadians) deliver 120V at 60Hz.

you need to explain what part the "ideal diode" is.

Now, if it was just a resistor and a battery in series, the answer is simple: (# of volts)/(# of ohms) = (# of amps = current).

Now if you can convert the diode and the power source into resistance, you add up all the resistance values and use the equation above.

As for the diode, connecting it in a reversed fashion (where no current flows through when the circuit is on) will make it become a low-value capacitor. BUT the capacitance is dependant on the voltage applied to the diode, and the diode itself.

You can get the impedance (What I like to call AC resistance) by using the formula 1/(2 * pi * capacitor value * frequency) where pi is "pie" and the frequency is the frequency the circuit is running at. (probably the same frequency as the AC generator). this "impedance" cannot be measured using an ohmmeter.

Let me tell ya, I learned the majority of my theory on the internet
:wink:

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-=: The best low-priced components to troubleshoot with are the speaker and the LED :=-

 

Of course law school teaches dishonesty and deception! What are lawyers for? Also remember most politicians have a degree in law. Need I say more?

 

Law school doesn't teach English very well either, does it? Man, for something in which how on earth do you survive? :lol: :lol:
Robert

 

But your theory is of limited use if you apply in indiscriminately.
An ideal diode has zero forward resistance, infinite reverse resistance, zero forward voltage drop, zero reverse leakage, and zero capacitance. In this case, frequency is irrelevant.
You can't make an ideal diode, and you can't buy one, but you can certainly theorize about one.

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Ron

 

There was only conduction when the diode was forward biased. Are you wondering about the Spice model? It's easy to model an ideal diode in Spice.

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Ron

 

I'm betting he doesn't use Internet shorthand in business communications.
Some of you guys are really prejudiced. I met a nice lawyer once. :lol:
Seriously, though, not all lawyers are bad. You don't even know this guy.

__________________
Ron

 

So we have
1) AC source: 16cos(400t)
2) DC source: 13V
3) Diode (ideal)
4) resistor: 2R

forget about resistor for now, concentrating on working out average voltage across resistor (this its a simple case of V = IR)

THe voltage (no diode) is:

V = 16cos(400t) + 13

So how to tackle this? Well easiest way it to find the points (in one full period) where the zero-crossing occurs.

For V = 0 =>
-13 = 16cos(400t)
arccos(-13/16) = 400*t

=> t = +/- 6.298ms

So integrate [16cos(400t) + 13] between +/-6.298m (and devide by the period) will give you the average voltage

Vavg = 0.2103864660/PERIOD


wt = 400t => w = 400
f = w/2pi => 2pi/w = 2pi/400
PERIOD = 0.015707963


Vavg = 13.39361841



Now with a 2R resistor puts the average amps at 13.39361841/2


Iavg = 6.696809205




As far as I see and quote
He came to this website to find the solution and he has.

not really an engineering question, more of a maths question but still, happy driving

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Nothing is impossible.
Once a problem is realised, the rest is just details

 

Ron,
It is not that I have prejudice for lawyers, I don't think he is one. It wasn't his 'net hand' that bothered me. (Lord knows mine sucks!) was I was only making fun of his misuse of the possesive, when he wanted the contraction you're. I would think a lawyer would write very well.

 

OMG, I missed that! And I am a compulsive proofreader! ops: :lol:

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Ron

 

Well, he now has the same answer from two different sources. But what if the nephew intended for the diode to point the other way? Styx, you're a good mathematician. I gave a simulated value (196.9mA). If you gave a theoretical one, it would have more credence.

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Ron