Theoretically value of R Load

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alishadevochka 18 years ago

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.

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Tim Wescott 18 years ago

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

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alishadevochka 18 years ago

Sorry, they are are.

formatting link

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alishadevochka 18 years ago

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...

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Helmut Sennewald 18 years ago

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

Hello,

The way to go is damn easy.

Write the formula for P(Rload).

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

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Helmut Sennewald 18 years ago

Hello,

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

Helmut

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Fred Bloggs 18 years ago

What's your question?

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alishadevochka 18 years ago

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

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alishadevochka 18 years ago

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

Oooh, thanks :-)

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Don Lancaster 18 years ago

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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

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Helmut Sennewald 18 years ago

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

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Fred Bloggs 18 years ago

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

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Theoretically value of R Load

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