Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

question on thevenin voltage

Status
Not open for further replies.
please have a look at the attached file.

using nodal analysis, find va and vb and vth=va-vb. rth=3r/2. just apply the thevenin method and solve your circuit.
REGARDS,
MEHTAB.
 
Last edited:
Firstly remove R1 then examine both sides separately.

Remember a current source has infinite resistance and a voltage source has zero impedance so in analysis you remove the current source for seeing the internal impedance and short out the voltage source. These are the fundamentals!

On the LHS you have a Norton (or current) equivalent circuit that by converting to a Thevenin will make life easier. So the LHS becomes a Thevenin (voltage source) of magnitude iR. This voltage source has a series impedance of R.

On the RHS using the voltage divider rule you have 1/2 of V with an equivalent impedance of two Rs in parallel i.e R/2 (Thevenin analysis).

Therefore if R1 is re-inserted, the current i1 through it from Ohm's law is

iR - V/2 (the differential voltage across R1 from both sources)
-----------------
(R + R1 + R/2) (the total resistance in the chain)

Mulitply top & bottom by two and

2iR - V
-----------
2R1 + 3R

View attachment 63682
 
Last edited:
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top