Theoretically value of R Load

I am stuck with the following question

Theoretically, for the network of Fig 11.7, what value of RL should result in maximum power to RL. I am doing Thevenin's theorem Validating the condition RL = Rth.

circuit is the series

Fig 11.7 is Eth 8v, R1 327, RL is the 1k (measure .998k ohm) pot from

0 to l

Thanks in advance.

Reply to
alishadevochka
Loading thread data ...

Boy, I've seen posts that assume that we all know the same thing, but this has to be some extreme limit.

Figure 11.7 in _what_? Are you asking us to do your homework for you?

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply to
Tim Wescott

Sorry, they are are.

formatting link
formatting link

Reply to
alishadevochka

I should've answer this question. I am not asking you or anyone to do my home. I am trying to understand this question. Cause I have no clue as to what I suppose to thing about when I am looking at the circuit or the graph...

Reply to
alishadevochka

schrieb im Newsbeitrag news: snipped-for-privacy@i13g2000prf.googlegroups.com...

Hello,

The way to go is damn easy.

  1. Write the formula for P(Rload).

  1. Differentiate it. dP(Rload)/dRload

3.Search the maximum. dP(Rload)/dRload=0 Calculate Rload from this formula.

You should be able to do this math.

Best regards, Helmut

Reply to
Helmut Sennewald

Hello,

Why has the calculated power the unit V in your worksheet? (22.47V ???)

Helmut

Reply to
Helmut Sennewald

What's your question?

Reply to
Fred Bloggs

Newsbeitragnews: snipped-for-privacy@i13g2000prf.googlegroups.com...

Formula for P Rload is

P max = I^2n * Rl /4

24.46mA^2 & .327 kohm / 4

48.910mW :-)

Cool thanks

Reply to
alishadevochka

Newsbeitragnews: snipped-for-privacy@j4g2000prf.googlegroups.com...

Oooh, thanks :-)

Reply to
alishadevochka

formatting link

--
Many thanks,

Don Lancaster                          voice phone: (928)428-4073
Synergetics   3860 West First Street   Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml   email: don@tinaja.com

Please visit my GURU's LAIR web site at http://www.tinaja.com
Reply to
Don Lancaster

schrieb im Newsbeitrag news: snipped-for-privacy@m37g2000prh.googlegroups.com...

Hello,

It seems you really need help. Vb=8V, R1=330

V=Vb*RL/(R1+RL)

P=V^2/RL

P=Vb^2*RL/(R1+RL)*RL/(R1+RL)/RL

P=Vb^2*RL/(R1+RL)^2

dP/dRL=Vb^2*(1/(R1+RL)^2-2*RL/(R1+RL)^3)

0=Vb^2*(1/(R1+RL)^2-2*RL/(R1+RL)^3)

0=1/(R1+RL)^2-2*RL/(R1+RL)^3

1/(R1+RL)^2=2*RL/(R1+RL)^3

1=2*RL/(R1+RL)

R1+RL=2*RL

R1=RL

RL=R1 for max. power RL=330Ohm

P=Vb^2*RL/(R1+RL)^2

Pmax = Vb^2*R1/(R1+R1)^2

Pmax=8*8*330/(660*660)W

Pmax=0.04848W

Best regards, Helmut

Reply to
Helmut Sennewald

There's a simple non-calculus method of showing this but it's what can be called cumbersome after you spell every damned thing out for a person on this level, sheesh: View in a fixed-width font such as Courier.

. . . . . + V1 - . . .---[R1]->>---. . | | + . | -I-> | . Es --- [Rx] Vx . - | . | | - . | | . .-------->>---' . . . . Some basic circuit formulas: . . . Es= V1 + Vx , sum of voltage drops is Es . . . Es . I = ------- , ckt I is Es divided by total R . R1 + Rx . 2 . Es . Pt = Es x I = ------ , total power delivered by Es to ckt . R1 + Rx . . . P1 = V1 x I . power dissipated in R1 . . . R1 . = Es x ------- x I . R1 + Rx . . . . Px = Vx x I , power dissipated in Rx . . Rx . = Es x ------- x I . R1 + Rx . . . Px Rx . then -- = -- . P1 R1 . . . also Pt= Es x I = ( V1 + Vx ) x I= V1 x I + Vx x I= P1 + Px . . . . Pt = P1 + Px . . 2 . Es . define Pto = -- = power delivered by Es when Rx = 0, . R1 . . . note that Pt ranges over 0

Reply to
Fred Bloggs

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.