# Thevenin

#### gbox

Joined Dec 29, 2015
42

I have found that I_1=3.81A and I_3=3.09A and I_2=0.72A and R_th=5.45Ω.
What do I need to do to find V_th?

#### WBahn

Joined Mar 31, 2012
29,885
You need to find the voltage between terminals a and b.

#### gbox

Joined Dec 29, 2015
42
so I need to calculate this loop:

But it is an open circuit, does the battery give voltage? does the 4\Omega resistor drop voltage?

#### crutschow

Joined Mar 14, 2008
34,074
so I need to calculate this loop:
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But it is an open circuit, does the battery give voltage? does the 4\Omega resistor drop voltage?
Yes, a battery has (gives) voltage, even when it is an open circuit.

The resistor will drop voltage only if there's current through it, i.e. if there's a load connected between A and B that draws current.

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#### WBahn

Joined Mar 31, 2012
29,885
so I need to calculate this loop:

But it is an open circuit, does the battery give voltage? does the 4\Omega resistor drop voltage?
This diagram has I2 as the current flowing downward in the 2 Ω resistor. But in your original diagram I2 was defined as the current flowing to the right through the adjacent 4 Ω resistor (the one to the left). You need to be consistent in your use of variables.

A battery is a voltage source and a voltage source maintains a specified voltage across it regardless of the current through it -- including a current of zero.

A resistor drops voltage across it in direct proportion to the current flowing through it -- if there is no current through it, then there is no voltage across it.

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#### WBahn

Joined Mar 31, 2012
29,885
Now, if the reason that you have I2 flowing downward in this diagram is because there is no current in the rightmost 4 Ω resistor and therefore I2 must also be the current flowing in the 2 Ω resistor, then that is correct and reasonable.