Finding Thevenin equivalent and maximum power transfer

[diffeqnoob]

Homework Statement

• Basically asking me to find the Thevenin/Norton's equivalent across a-b.
• Then figure out the maximum power transferred
• Then figure out max power delivered

Homework Equations

• Maximum power is transferred when $$R_{l} = R_{th}$$ .
  [*]Max power delivered is $$\frac{V_2}{4R_{th}}$$
  

The Attempt at a Solution

I honestly don't know where to start this. I know Thevenin equivalent requires me to break this down into a voltage source, one resistor in series, but I really have no clue where to start. I tried using nodal analysis but this left me with nothing.

--My poor attempt at nodal analysis--

Node1:
$$-IR = 4$$

$$I = \frac{-4}{R}$$

Node2:
$$5I_1$$ = -3$$I_1$$

--sub $$\frac{-4}{R}$$ for $$I_1$$--$$5\frac{-4}{R} = -3\frac{-4}{R}$$

$$\frac{-20}{R}= \frac{12}{R}$$

$$-20 = 12$$
?

A push in the right direction would be greatly appreciated.

 

[dashkin111]

You will need to do nodal analysis twice. Once with the independent sources on, once with them off (the later to find the Thevenin/Norton equivalent resistance).

Set the bottom to the ground node and the top is $$V_{th}$$. This will yield one equation with one unknown- a straightforward solution.


To find $$R_{th}$$ turn off the independent source (the 4A current source), connect a test source between A and B and find the equivalent resistance from there.

 

[diffeqnoob]

Thank you very much! I will try this tomorrow morning and post my results. Again, thanks for the response.

 

[dashkin111]

No problem, it was good review for me as I have an exam on it monday :P

 

[diffeqnoob]

Okay. I think I got it.

Now, using KCL at the node above the dependent current source I have:

$$0 = 4 - 3_i_1 - i_1$$
$$4 = 4_i_1$$
$$\ldots i_1 = 1$$ amp

$$V_{oc} = 5 V$$ $$\ldots$$ $$V_T = 5 V$$
$$I_{sc} = 4 A$$

$$R_{th} = \frac{V_{oc}}{I_{sc}}$$

$$R_{th} = \frac{5}{4} \Omega$$

So, a voltage source of 5V in series with a resistor of $$\frac{5}{4} \Omega$$. Now to figure out the max power delivered and max power transferred.