Linear Circuits Thevenins Theorem question

[careless25]

Hi,

New member here!

Ok, this is a textbook question which I need help with as I am self teaching myself Linear Circuits. I have an exam in a week and I wanted to clear out some questions I had.

Homework Statement

textbook question: https://docs.google.com/leaf?id=0By2PVt4Tdl6VZjNmMGRmODgtOWVmYy00MTUxLWExNjItZmQ4MTMxMzYyZDAy&hl=en_GB

Find Vo using Thévenin’s theorem

Homework Equations

Mesh Analysis:

I1 = 8mA (8mA current source loop)
I2 = 2mA (2mA current source loop)

Loop I3 : -I1(3kΩ) - I2(6kΩ) + I3(15kΩ) = -12V (voltage source loop)
I3 = 36/15 = 12/5 mA

R_th = 18/5 kΩ + 2kΩ

The Attempt at a Solution

Since we want to find Voltage across the 1kΩ resistor using Thevenin's theorem, I removed the resistor from the circuit.
Now to find the voltage across the Open terminals I would have to simplify the circuit. Where I am stuck is how I can simplify the rest of the circuit so that I have Voltage and R_th.
The current sources are baffling me, I don't know how to simplify those.

I tried using Mesh analysis because i have 2 of the 3 currents but I don't know how to use those currents to find a voltage across the terminals.

Any help would be appreciated.

Thanks,

C25

 

[gneill]

With your calculated I3 you should be able to find the voltage drop across the 6kΩ resistor which is in series with the 12V supply. The 2mA current source forces the potential drop across the 2kΩ resistor. Add a touch of KVL and you have your Thevenin voltage.

 

[MisterX]

Here are some things to know.

The Thevenin and Norton resistances are equivalent.

The short circuit load current is equal to the Norton current.

The Thevenin and Norton resistances are equal to the Thevenin voltage (a.k.a. short circuit load voltage) divided by the Norton current.