/!\Thevenin Theorem prob. T___T /!\

!!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 <<<< 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

!!Miss.EE!!

Joined Oct 17, 2007
38
http://img84.imageshack.us/img84/7082/121lg9.png

click the link above !!

Or Regards !!Miss.EE!!

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