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Thevenin / Superposition

198
0
Never mind, did superposition incorrectly.

1. Homework Statement

http://img258.imageshack.us/img258/9206/jesus2gn2.th.png [Broken]

Find Vload using superposition and thevenin's equivalent

2. Homework Equations



3. The Attempt at a Solution

Well for network on the left of Rload, I found Vth = 5 V and Rth = 15.1 kohms
For the network on the right of Rload, I found Vth = 7.5 V and Rth = 8150 ohms.

So using superposition I would do:
1) 5 V * 10k / (10k+15.1k) = 1.99 V on Rload
2) 7.5 V * 10k / (10k + 8150) = 4.13 V
Adding those I get 6.12 V.

However, when I simulate this in PSPICE I get 4.33 V.
I also checked each side and I get 1.99 V on the left (when I disconnect the right) and I get 4.13 V on the right (when I disconnect the left).

What am I doing wrong, I can't even remotely see where that 4.33 V comes from.
 
Last edited by a moderator:

Answers and Replies

454
0
When you calculate the voltage on the load due to one voltage source, you can't just remove the others, but need to replace them with their internal resistance.

for 1) you'd get 5 v * (10K // 8150) / ((10K // 8150) + 15.1k)
(A // B) is the value of the resistance of A parallel with B, or AB/(A+B)
 

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