 |
|
 |
 |
 |
 |
Forum: Electrical & Computer Engineering :: Electric Circuits :: Thevenin and Norton  | |||||||||
| |||||||||
| Author | Message | ||||||||
| aac13 | #3 | Sat 4 Apr 2009 4:07:00 PM | ||||||||||
Student Joined Wed 21 Jan 2009 Posts 151 |
Find the Thévenin equivalent with respect to the terminals a, b in the circuit shown. 
|
|||||||||||
|
|
|||||||||||
| aac13 | Sat 4 Apr 2009 4:08:00 PM | |||||||||||
|
Student Joined Wed 21 Jan 2009 Posts 151 |
IT = (vT – v1)/20 + (vT – 40i∆)/60
(v1 – 40v∆)/16 + v1/80 + (v1 – vT)/20 = 0
i∆ = - v1/80 = -0.32vT /80 40i∆ = - 0.16vT
iT = (vT – 0.32vT)/20 + (vT + 0.16vT)/60
= ( (0.68 vT)(3) + 1.16vT ) /60
= ( (2.04 + 1.16)vT )/60 = 3.20vT/60
VT /iT = 60/3.2 = 18.75 Ω
è RTh = 18.75 Ω |
|||||||||||
|
|
|||||||||||
| 1 | ||||||||||||
3/5
 |
|
 |