/!\Thevenin Theorem prob. T___T /!\

Thread Starter

!!Miss.EE!!

Joined Oct 17, 2007
38
Hi

here I am with new problem I faced.....this time as quize >>Make up Quize and what I've done .... I don't think that much <<<< :p

Hoever , the picture is below and the question is ubone it ......


what Ive done is saying that the current through R1 and R4 is 1 mA when I openen the circuit <<< Removing RL>>>, and the current through R2 and R3 is (Vx/4000) and I solved the problem But I don't think what I've done is correct so please any one can explaine ..... Regards to all !!Miss.EE!!:)

 

hgmjr

Joined Jan 28, 2005
9,029
Your attachment appears to have failed to make it into your post. At least I can't see it with my browser (IE7).

hgmjr
 

chuckey

Joined Jun 4, 2007
75
1. You have 2 current sources lets call it (Vx/4) +1 mA or I
2. Thevenim, he say remove the load and see how much voltage you have. To do this you have to work out the current division through the two resistor chains , from this you can then work out the voltage drop at each end of Rl.
chain 1 = R1 + R4 other, chain 2 R2 +R3
current through 1 = I X (R1 + R4)/R1+R2+R3+R4
current through 2 = I X (R2 + R3)/R1+R2+R3+R4
therefore voltage at top end of R4 = I X R4(R1 + R4)/R1+R2+R3+R4.........1
therefore voltage at top end of R3 = I X R3(R2 + R3)/R1=R2+R3+R4.........2
therefore voltage across Rl terminals = eqt.1- eqt. 2
Internal impedance of Ts equiv =
Two paths in series (R2,R3 and R1,R4)
Equiv of R2&R3 = R2 X R3/R2+R3........................3
equiv of R1&R4 = R1 X R4/R1+R4........................4
Therefore total internal impedance = eqt.3 + eqt.4
Best of luck Frank -better check my maths I'm a bit (a lot) rusty
 
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