Hey guys, here are some questions regarding Thevenin's Theorem For this question. it is stated that no load resistance is connected, not really sure what does this mean though. I'm supposed to find Vth for this question. Vs = 6v, Is = 2mA, R1 = 2 kΩ, R2 = 4 kΩ, R3 = 6 kΩ First, I break the network at the 6k load. Ok so for this question, I started with the mesh equations with the top left mesh being I1 in the clockwise direction and I2 the bottom mesh also going in a clockwise direction. -6 + 4K(I1) +2k( I1-I2) = 0 and I obtained I1 = 5/3 mA I2 = 2 x 10^-3 Using KVL, Voc is 4k(I1) + 2k(I2) = 4k(5/3 x 10^-3) + 2k(2 x 10^-3) = 32/3 V = Vth Rth = (2k//4k) + 2k = 10/3 kΩ Would this be correct? For this question. it is stated that no load resistance is connected, not really sure what does this mean though. I'm required to find the Rth for this question. R1 = 9.5 kΩ: R2 = 8.0 kΩ VS1 = 3 V : VS2 = 3 V : IS = 50 mA This one I honestly have no idea at all so I hope you guys can help me. Thanks in advanced, sorry if its too lengthy..
No load resistance means that there is no load (nothing that draws current) between the A and B terminals of the circuit. If you disable the sources and calculate the effective resistance across the terminals you et 2142 ohm, which is not the answer you got. So somewhere you went wrong. By the looks of it you forgot the R3 resistor when you calculated Rth which is in parallel with the answer you did calculate.
Arr, it's not that I forgot it... I broke the circuit at the 6k load. This question only requires the calculation of Vth, not Vo. But thanks for your info. Still need help from you guys, heh.
I think the question asks for the thevenin's equivalent as seen by the load if it is connected in parallel with R3. so you have three meshes and solve for the two currents(one being the same value as Is) and use this to find the voltage across R3 which is the Vth. For Rth shorting the voltage source and opening the current source we have ((R1||R2 ) + R1) ||R3
For the second question again if the load is to be connected parallel to the terminals R2(marked as Vo) then open circuiting the current sources and shorting the voltage sources, we have two resistors in parallel above three resistors (R1,R2,R2) connected IN PARALLEL. Now you dont need to take into account the upper two resistors .You need to calculate the equivalent resistance of R1||R2||R2 which will give the Rth. For calculating Vth you can use node method with centre node being the reference
I calculated (R1||R2 ) + R1) as Rth for Q1 and still got it wrong...it sorts of explains it I guess. Had the same problem for Q2 by calculating Rth as (R1||R2). Can you explain again why must the load resistor be considered as part of the Rth? Thanks, cause I still don't really understand. >.<
There is no load resistor in the circuit. R3 is part of the circuit and that's why it's taken into calculation of Rth. A load resistor is usually always called Rl or Rload. If it were in the circuit there would be a fifth resistor, connected directly between the A and B terminals. Hope that helped
Actually I'm still having some problems with Q1, would taking the Voc I calculated in post 1 to caltulate Vth yield the correct answer? Vo = Voc(R3/(R3+Rth)) = Vth? Where Voc = 32/3V and Rth = ((R1||R2 ) + R1) ||R3
This will work if you remember that Voc was calculated with R3 missing, so you need to use a value for Rth that is also calculated with R3 missing. In other words, let Rth = ((R1||R2 ) + R1) and you'll get the right answer which is Vth = 48/7