http://gyazo.com/7e1d8f17da87aba122d37d93c221a9ca (Screenshot of the problem)

 

Homework?

(Show us a screenshot of your attempted solution).

 

Homework?

(Show us a screenshot of your attempted solution).

 

Hi, to me your solution is wrong .
For Vd = 0.7V diode current is equal to
Id = (4*0.7V + 5*0.7^2) = 2.8 + 2.45 = 5.25mA
And resistor current
Ir = (5V - 0.7V)/200Ω = 21.5mA
So Id is not equal to Ir. and this means that your solution is wrong.
And the correct answer is Vd ≈ 1.51V because this time
Id ≈ 4*1.51V + 5*1.51V^2 ≈ 17.4405mA and Ir = (5V - 1.51V)/200Ω ≈ 17.45mA
And since Id≈Ir we have a correct solution.
Additional I plot Id vs Vd and load line

 

where is the 1.51v coming from?

 

Well I simply solve this
(5V - Vd)/200Ω = (4*Vd + 5*Vd^2)/1000
http://www.wolframalpha.com/input/?i=(5+-+V)/200+=+(4*V+++5*V^2)/1000+

 

Or Let LTSpice solve that equation. The B1 behavioral current source has the same I/V behavior as the diode...

 

Hi guys, didnt want to create a new thread.

Any help would be appreciated with this problem

http://gyazo.com/797d994c1b5617555d9565d072aebc76

 

http://electronics.stackexchange.com/questions/72130/half-wave-rectifier-and-rms-value
http://forum.allaboutcircuits.com/t...ll-wave-and-half-wave-rectifier-signal.62170/

 

230 V * 1/5 = 46Vrms on the secondary

Output pk voltage will be 46 * sqrt(2)/pi = 20.70V = DC

RMS = 20.70V/2= 10.35V

P = I^2 * R = Vrms^2/R

P = 10.35^2/10 = 10.71W


Does this look correct to you guys?